RS Aggarwal Solutions for Class 11 Chapter 28 Differentiation

Question 1. Differentiate the following functions:
(i) \( x^{-3} \)
(ii) \( \sqrt[3]{x} \)
Answer:
(i) \( x^{-3} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}x^{-3} = -3x^{-3-1} = -3x^{-4} \)

(ii) \( \sqrt[3]{x} = x^{\frac{1}{3}} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}x^{\frac{1}{3}} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} \)
In simple words: To find the derivative of a power of x, bring down the exponent and reduce it by one. Apply this rule to each term separately.

Exam Tip: Always convert roots and fractions to exponent form before applying the power rule - this makes the calculation straightforward and less error-prone.

Question 2. Differentiate the following functions:
(i) \( \frac{1}{x} \)
(ii) \( \frac{1}{\sqrt{x}} \)
(iii) \( \frac{1}{\sqrt[3]{x}} \)
Answer:
(i) \( \frac{1}{x} = x^{-1} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}x^{-1} = -1x^{-1-1} = -x^{-2} \)

(ii) \( \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}x^{-\frac{1}{2}} = -\frac{1}{2}x^{-\frac{1}{2}-1} = -\frac{1}{2}x^{-\frac{3}{2}} \)

(iii) \( \frac{1}{\sqrt[3]{x}} = x^{-\frac{1}{3}} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}x^{-\frac{1}{3}} = -\frac{1}{3}x^{-\frac{1}{3}-1} = -\frac{1}{3}x^{-\frac{4}{3}} \)
In simple words: Rewrite each fraction using negative exponents, then apply the power rule. The negative exponent comes down, and the exponent decreases by one.

Exam Tip: Convert fractional and reciprocal forms to exponential notation first - this prevents common errors and makes the pattern clearer.

Question 3. Differentiate the following functions:
(i) \( 3x^{-5} \)
(ii) \( \frac{1}{5x} \)
(iii) \( 6.\sqrt[3]{x^2} \)
Answer:
(i) \( 3x^{-5} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}3x^{-5} = 3(-5)x^{-5-1} = -15x^{-6} \)

(ii) \( \frac{1}{5x} = \frac{1}{5}x^{-1} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{1}{5} \frac{d}{dx}x^{-1} = \frac{1}{5} \times (-1)x^{-1-1} = -\frac{1}{5}x^{-2} \)

(iii) \( 6.\sqrt[3]{x^2} = 6x^{\frac{2}{3}} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}6x^{\frac{2}{3}} = 6 \times \frac{2}{3}x^{\frac{2}{3}-1} = 4x^{-\frac{1}{3}} \)
In simple words: For each term, multiply the coefficient by the exponent, then reduce the exponent by one. This works the same way whether the exponent is positive, negative, or fractional.

Exam Tip: Handle coefficients and exponents separately - multiply the coefficient by the exponent, subtract one from the exponent, and keep the variable.

Question 4. Differentiate the following functions:
(i) \( 6x^5 + 4x^3 - 3x^2 + 2x - 7 \)
(ii) \( 5x^{-\frac{3}{2}} + \frac{4}{\sqrt{x}} + \sqrt{x} - \frac{7}{x} \)
(iii) \( ax^3 + bx^2 + cx + d, \) where a, b, c, d are constants
Answer:
(i) \( 6x^5 + 4x^3 - 3x^2 + 2x - 7 \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}(6x^5 + 4x^3 - 3x^2 + 2x - 7) = 30x^{5-1} + 12x^{3-1} - 6x^{2-1} + 2x^{1-1} + 0 \)
\( = 30x^4 + 12x^2 - 6x + 2 \)

(ii) \( 5x^{-\frac{3}{2}} + \frac{4}{\sqrt{x}} + \sqrt{x} - \frac{7}{x} \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}(5x^{-\frac{3}{2}} + 4x^{-\frac{1}{2}} + x^{\frac{1}{2}} - 7x^{-1}) \)
\( = 5 \times (-\frac{3}{2})x^{-\frac{3}{2}-1} + 4 \times (-\frac{1}{2})x^{-\frac{1}{2}-1} + \frac{1}{2}x^{\frac{1}{2}-1} - 7 \times (-1)x^{-1-1} \)
\( = -\frac{15}{2}x^{-\frac{5}{2}} - 2x^{-\frac{3}{2}} + \frac{1}{2}x^{-\frac{1}{2}} + 7x^{-2} \)

(iii) \( ax^3 + bx^2 + cx + d, \) where a, b, c, d are constants
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \)

Differentiating with respect to x,
\( \frac{d}{dx}(ax^3 + bx^2 + cx + d) = 3ax^{3-1} + 2bx^{2-1} + cx^{1-1} + d \times 0 \)
\( = 3ax^2 + 2bx + c \)
In simple words: Differentiate each term separately. For each term, bring down the exponent as a coefficient and reduce the exponent by one. Constants vanish when differentiated.

Exam Tip: Work term by term and apply the power rule consistently - constants become zero, linear terms become constants, and higher powers drop by one exponent level.

Question 5. Differentiate the following functions:
(i) \( 4x^3 + 3.2^x + 6.\sqrt[8]{x^{-4}} + 5\cot x \)
(ii) \( \frac{x}{3} - \frac{3}{x} + \sqrt{x} - \frac{1}{\sqrt{x}} + x^2 - 2^x + 6x^{-\frac{2}{3}} - \frac{2}{3}x^6 \)
Answer:
(i) \( 4x^3 + 3.2^x + 6.x^{-\frac{1}{2}} + 5\cot x \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \), \( \frac{d}{dx}\cot x = -\csc^2 x \), and \( \frac{d}{dx}a^x = \log_n(a) \times a^x \)

Differentiating with respect to x,
\( \frac{d}{dx}(4x^3 + 3.2^x + 6x^{-\frac{1}{2}} + 5\cot x) \)
\( = 4 \times 3x^{3-1} + 3.\log_n(2).2^x + 6 \times (-\frac{1}{2})x^{-\frac{1}{2}-1} + 5 \times (-\csc^2 x) \)
\( = 12x^2 + 3.\log_n(2).2^x - 3x^{-\frac{3}{2}} - 5\csc^2 x \)

(ii) \( \frac{x}{3} - 3x^{-1} + x^{\frac{1}{2}} - x^{-\frac{1}{2}} + x^2 - 2^x + 6x^{-\frac{2}{3}} - \frac{2}{3}x^6 \)
Using the formula \( \frac{d}{dx}x^n = nx^{n-1} \) and \( \frac{d}{dx}a^x = \log_n(a) \times a^x \)

Differentiating with respect to x,
\( \frac{d}{dx}(\frac{x}{3} - 3x^{-1} + x^{\frac{1}{2}} - x^{-\frac{1}{2}} + x^2 - 2^x + 6x^{-\frac{2}{3}} - \frac{2}{3}x^6) \)
\( = \frac{1}{3} - 3 \times (-1)x^{-1-1} + \frac{1}{2}x^{\frac{1}{2}-1} - (-\frac{1}{2})x^{-\frac{1}{2}-1} + 2x^{2-1} - \log_n(2).2^x + 6 \times (-\frac{2}{3})x^{-\frac{2}{3}-1} - \frac{2}{3} \times 6x^{6-1} \)
\( = \frac{1}{3} + 3x^{-2} + \frac{1}{2}x^{-\frac{1}{2}} + \frac{1}{2}x^{-\frac{3}{2}} + 2x - \log_n(2).2^x - 4x^{-\frac{5}{3}} - 4x^5 \)
In simple words: Convert all terms to exponential form, then apply the power rule to each. Trigonometric and exponential functions have their own specific differentiation rules that you apply separately.

Exam Tip: Always rewrite roots and reciprocals using exponents before differentiating - this ensures you can apply the power rule uniformly across all polynomial terms.

Question 6. Differentiate the following functions:
(i) \( 4\cot x - \frac{1}{2}\cos x + \frac{2}{\cos x} - \frac{3}{\sin x} + \frac{6\cot x}{\cos ec x} + 9 \)
(ii) \( -5\tan x + 4\tan x \cos x - 3\cot x \sec x + 2\sec x - 13 \)
Answer: Formulae:
\( \frac{d}{dx}\cot x = -\csc^2 x \)
\( \frac{d}{dx}\cos x = -\sin x \)
\( \frac{d}{dx}\sec x = \sec x \tan x \)
\( \frac{d}{dx}\cos ec x = -\cos ec x \cot x \)
\( \frac{d}{dx}\tan x = \sec^2 x \)
\( \frac{d}{dx}\sin x = \cos x \)
\( \frac{d}{dx}k = 0, \) k is constant

(i) \( 4\cot x - \frac{1}{2}\cos x + 2\sec x - 3\cos ec x + 6\cos x + 9 \)

Differentiating with respect to x,
\( \frac{d}{dx}(4\cot x - \frac{1}{2}\cos x + 2\sec x - 3\cos ec x + 6\cos x + 9) \)
\( = 4(-\csc^2 x) - \frac{1}{2}(-\sin x) + 2\sec x \tan x - 3(-\cos ec x \cot x) + 6(-\sin x) + 0 \)
\( = -4\csc^2 x + \frac{1}{2}\sin x + 2\sec x \tan x + 3\cos ec x \cot x - 6\sin x \)

(ii) \( -5\tan x + 4\sin x - 3\cos ec x + 2\sec x - 13 \)

Differentiating with respect to x,
\( \frac{d}{dx}(-5\tan x + 4\sin x - 3\cos ec x + 2\sec x - 13) \)
\( = -5\sec^2 x + 4\cos x - 3(-\cos ec x \cot x) + 2\sec x \tan x - 0 \)
\( = -5\sec^2 x + 4\cos x + 3\cos ec x \cot x + 2\sec x \tan x \)
In simple words: Each trigonometric function has its own derivative formula. Apply the rule for each function separately, then combine the results. Constant terms vanish during differentiation.

Exam Tip: Memorize the derivatives of all six trigonometric functions and watch the signs carefully - especially the negative signs in cosine, cosecant, and cotangent derivatives.

Question 7. Differentiate the following functions:
(i) \( (2x + 3)(3x - 5) \)
(ii) \( x(1 + x)^3 \)
(iii) \( (\sqrt{x} + \frac{1}{x})(x - \frac{1}{\sqrt{x}}) \)
(iv) \( (x - \frac{1}{x})^2 \)
(v) \( (x^2 - \frac{1}{x^2})^3 \)
(vi) \( (2x^2 + 5x - 1)(x - 3) \)
Answer: Formula:
\( \frac{d}{dx}f(g(x)) = \frac{d}{dg}f(g) \frac{d}{dx}g \)

Chain Rule:
\( \frac{d}{dx}(uv) = u\frac{d}{dx}v + v\frac{d}{dx}u \)

Where u and v are the functions of x.

(i) \( (2x + 3)(3x - 5) \)

Applying Chain Rule:
Here, u = 2x + 3
v = 3x - 5

\( \frac{d}{dx}(2x + 3)(3x - 5) = (2x + 3)\frac{d}{dx}(3x - 5) + (3x - 5)\frac{d}{dx}(2x + 3) \)
\( = (2x + 3)(3) + (3x - 5)(2) \)
\( = 6x + 9 + 6x - 10 \)
\( = 12x - 1 \)

(ii) \( x(1 + x)^3 \)

Applying Chain Rule:
Here, u = x
v = (1 + x)^3

\( \frac{d}{dx}x(1 + x)^3 = x\frac{d}{dx}(1 + x)^3 + (1 + x)^3\frac{d}{dx}(x) \)
\( = x \times 3(1 + x)^2 + (1 + x)^3(1) \)
\( = (1 + x)^2(3x + x + 1) \)
\( = (1 + x)^2(4x + 1) \)

(iii) \( (x^{\frac{1}{2}} + x^{-1})(x - x^{-\frac{1}{2}}) \)

Applying Chain Rule:
Here, u = (x^{\frac{1}{2}} + x^{-1})
v = (x - x^{-\frac{1}{2}}) \)

\( \frac{d}{dx}(x^{\frac{1}{2}} + x^{-1})(x - x^{-\frac{1}{2}}) \)
\( = (x^{\frac{1}{2}} + x^{-1})\frac{d}{dx}(x - x^{-\frac{1}{2}}) + (x - x^{-\frac{1}{2}})\frac{d}{dx}(x^{\frac{1}{2}} + x^{-1}) \)
\( = (x^{\frac{1}{2}} + x^{-1})(1 + \frac{1}{2}x^{-\frac{3}{2}}) + (x - x^{-\frac{1}{2}})(\frac{1}{2}x^{-\frac{1}{2}} - x^{-2}) \)
\( = x^{\frac{1}{2}} + x^{-1} + \frac{1}{2}x^{-1} + \frac{1}{2}x^{-\frac{5}{2}} + \frac{1}{2}x^{\frac{1}{2}} - x^{-1} - \frac{1}{2}x^{-1} + x^{-\frac{5}{2}} \)
\( = \frac{3}{2}x^{\frac{1}{2}} + \frac{3}{2}x^{-\frac{5}{2}} \)

(iv) \( (x - \frac{1}{x})^2 \)

Using composite differentiation: \( \frac{d}{dx}f(g(x)) = \frac{d}{dg}f(g)\frac{d}{dx}g \)

Here, f(g) = g^2, g(x) = x - \frac{1}{x}

\( \frac{d}{dx}(x - \frac{1}{x})^2 = 2g(1 + \frac{1}{x^2}) \)
\( = 2(x - \frac{1}{x})(1 + \frac{1}{x^2}) \)
\( = 2(x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^3}) \)
\( = 2(x + \frac{1}{x^3}) \)

(v) \( (x^2 - \frac{1}{x^2})^3 \)

Using composite differentiation: \( \frac{d}{dx}f(g(x)) = \frac{d}{dg}f(g)\frac{d}{dx}g \)

Here, f(g) = g^3, g(x) = x^2 - \frac{1}{x^2}

\( \frac{d}{dx}(x^2 - \frac{1}{x^2})^3 = 3g^2(2x - \frac{2}{x^3}) \)
\( = 3(x^2 - \frac{1}{x^2})^2(2x - \frac{2}{x^3}) \)

(vi) \( (2x^2 + 5x - 1)(x - 3) \)

Applying Chain Rule:
Here, u = (2x^2 + 5x - 1)
v = (x - 3)

\( \frac{d}{dx}(2x^2 + 5x - 1)(x - 3) \)
\( = (2x^2 + 5x - 1)\frac{d}{dx}(x - 3) + (x - 3)\frac{d}{dx}(2x^2 + 5x - 1) \)
\( = (2x^2 + 5x - 1) \times 1 + (x - 3)(4x + 5) \)
\( = 2x^2 + 5x - 1 + 4x^2 - 7x - 15 \)
\( = 6x^2 - 2x - 16 \)
In simple words: When you have a product of two functions, use the product rule: the first function times the derivative of the second, plus the second function times the derivative of the first. For composite functions, apply the chain rule by bringing down the exponent and multiplying by the derivative of what is inside.

Exam Tip: Always identify whether you need the product rule (for multiplied functions) or chain rule (for nested/composite functions). Check your brackets and signs carefully when expanding the final answer.

Question 8. Differentiate the following functions:
(i) \( \frac{3x^2 + 4x - 5}{x} \)
(ii) \( \frac{(x^3 + 1)(x - 2)}{x^2} \)

Question 9. (i) If \( y = 6x^5 - 4x^4 - 2x^2 + 5x - 9 \), find \( \frac{dy}{dx} \) at \( x = -1 \).
Answer: Differentiating with respect to x,
\[ \frac{d}{dx}(6x^5 - 4x^4 - 2x^2 + 5x - 9) = 30x^4 - 16x^3 - 4x + 5 \]

Substituting \( x = -1 \):
\[ \left(\frac{dy}{dx}\right)_{x=-1} = 30(-1)^4 - 16(-1)^3 - 4(-1) + 5 \]
\[ = 30 + 16 + 4 + 5 = 55 \]
In simple words: Take the derivative by applying the power rule to each term. Then plug in x equals negative one and calculate the result.

Exam Tip: Apply the power rule term by term and ensure signs are handled correctly when substituting negative values.

Question 9. (ii) If \( y = (\sin x + \tan x) \), find \( \frac{dy}{dx} \) at \( x = \frac{\pi}{3} \).
Answer: Differentiating with respect to x,
\[ \frac{d}{dx}(\sin x + \tan x) = \cos x + \sec^2 x \]

Substituting \( x = \frac{\pi}{3} \):
\[ \left(\frac{dy}{dx}\right)_{x=\pi/3} = \cos \frac{\pi}{3} + \sec^2 \frac{\pi}{3} = \frac{1}{2} + 4 = \frac{9}{2} \]
In simple words: The derivative of sine is cosine, and the derivative of tangent is secant squared. Substitute the angle and evaluate using known trigonometric values.

Exam Tip: Memorise the derivatives of all trigonometric functions and the standard values at common angles like π/3.

Question 9. (iii) If \( y = \frac{2 - 3\cos x}{\sin x} \), find \( \frac{dy}{dx} \) at \( x = \frac{\pi}{4} \).
Answer: Differentiating with respect to x,
\[ \frac{d}{dx}(2\cos x - 3\cot x) = 2(- \csc x \cot x) - 3(- \csc^2 x) \]
\[ = -2\csc x \cot x + 3\csc^2 x \]

Substituting \( x = \frac{\pi}{4} \):
\[ \left(\frac{dy}{dx}\right)_{x=\pi/4} = -2\csc \frac{\pi}{4} \cot \frac{\pi}{4} + 3\csc^2 \frac{\pi}{4} = -2 \times \sqrt{2} \times 1 + 3 \times 2 = 6 - 2\sqrt{2} \]
In simple words: Apply the quotient rule or simplify the fraction first, then differentiate using trigonometric derivative formulas. Finally, evaluate at the given angle.

Exam Tip: Simplifying the expression before differentiating often makes the calculation easier and reduces the chance of errors.

Question 10. If \( y = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) \), show that \( 2x \frac{dy}{dx} + y = 2\sqrt{x} \).
Answer: We need to demonstrate:
\[ 2x \frac{dy}{dx} + y = 2\sqrt{x} \]

Differentiating y with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \]

Now, calculating the left-hand side:
\[ \text{LHS} = 2x \cdot \left(\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}\right) + \sqrt{x} + \frac{1}{\sqrt{x}} \]
\[ = \sqrt{x} - \frac{1}{\sqrt{x}} + \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{x} = \text{RHS} \]
In simple words: Find the derivative, then substitute both the derivative and the original function into the given equation. If both sides match, the relationship is verified.

Exam Tip: When asked to "show that," always finish by writing "LHS = RHS" to confirm the identity holds.

Question 11. If \( y = \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) \), prove that \( (2xy)\left(\frac{dy}{dx}\right) = \left(\frac{x}{a} - \frac{a}{x}\right) \).
Answer: We need to demonstrate:
\[ (2xy)\left(\frac{dy}{dx}\right) = \left(\frac{x}{a} - \frac{a}{x}\right) \]

Differentiating y with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) = \frac{1}{2\sqrt{ax}} - \frac{\sqrt{a}}{2x^{3/2}} \]

Now, calculating the left-hand side:
\[ \text{LHS} = 2x \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) \left(\frac{1}{2\sqrt{ax}} - \frac{\sqrt{a}}{2x^{3/2}}\right) \]
\[ = \left(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\right) \left(\sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}\right) \]
\[ = \frac{x}{a} - \frac{a}{x} = \text{RHS} \]
In simple words: Differentiate the given function, then expand the product by multiplying the derivative with the original function and the coefficient. The result simplifies to match the right-hand side.

Exam Tip: Watch for algebraic patterns like difference of squares - they often simplify the proof significantly.

Question 12. If \( y = \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}} \), find \( \frac{dy}{dx} \).
Answer:

Using the double angle formulas:
\[ \cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x \]
\[ \therefore 1 + \cos 2x = 2\cos^2 x \]
\[ 1 - \cos 2x = 2\sin^2 x \]

Substituting these:
\[ y = \sqrt{\frac{2\cos^2 x}{2\sin^2 x}} = \sqrt{\cot^2 x} = \cot x \]

Differentiating y with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx}(\cot x) = -\csc^2 x \]
In simple words: First simplify the expression using double angle identities to get it into a simpler form. Then differentiate the simplified result.

Exam Tip: Always look for opportunities to simplify using trigonometric identities before differentiating - it reduces computation and errors.

Question 13. If \( y = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \), find \( \frac{dy}{dx} \).
Answer:

Using the half-angle formula:
\[ \cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \]

Therefore:
\[ y = \cos x \]

Differentiating y with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx}(\cos x) = -\sin x \]
In simple words: Recognize that the given expression matches a standard half-angle identity for cosine. Once you simplify to y equals cosine x, the derivative is straightforward.

Exam Tip: Familiarity with half-angle and double-angle formulas is essential - they frequently appear in differentiation problems and dramatically simplify solutions.

Question 1. (Exercise 28B) Find the derivative of each of the following from the first principle: \( (ax + b) \)
Answer: Let \( f(x) = ax + b \)

We need to find the derivative of f(x), that is, \( f'(x) \)

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots(i) \]

\( f(x) = ax + b \)
\[ f(x + h) = a(x + h) + b = ax + ah + b \]

Putting values in (i), we get:
\[ f'(x) = \lim_{h \to 0} \frac{(ax + ah + b) - (ax + b)}{h} = \lim_{h \to 0} \frac{ah}{h} = \lim_{h \to 0} a = a \]

Therefore, \( f'(x) = a \)
In simple words: Apply the first principle formula by substituting the function at x plus h and at x. Simplify the difference, divide by h, and take the limit as h approaches zero.

Exam Tip: The first principle is the foundational method - always cancel h from numerator and denominator before taking the limit.

Question 2. (Exercise 28B) Find the derivative of each of the following from the first principle: \( \left(ax^2 + \frac{b}{x}\right) \)
Answer: Let \( f(x) = ax^2 + \frac{b}{x} \)

We need to find the derivative of f(x), that is, \( f'(x) \)

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots(i) \]

\[ f(x) = ax^2 + \frac{b}{x} \]
\[ f(x + h) = a(x + h)^2 + \frac{b}{(x + h)} \]

Putting values in (i), we get:
\[ f'(x) = \lim_{h \to 0} \frac{\left[a(x + h)^2 + \frac{b}{(x + h)}\right] - \left[ax^2 + \frac{b}{x}\right]}{h} \]
\[ = \lim_{h \to 0} \frac{a[(x + h)^2 - x^2] + b\left[\frac{1}{x + h} - \frac{1}{x}\right]}{h} \]
\[ = \lim_{h \to 0} \frac{a[h^2 + 2xh] + b\left[\frac{x - (x + h)}{x(x + h)}\right]}{h} \]
\[ = \lim_{h \to 0} \frac{a[h^2 + 2xh] + b\left[\frac{-h}{x(x + h)}\right]}{h} \]
\[ = \lim_{h \to 0} \left[a(h + 2x) - \frac{b}{x(x + h)}\right] \]

Putting h = 0, we get:
\[ f'(x) = a(0 + 2x) - \frac{b}{x(x + 0)} = 2ax - \frac{b}{x^2} \]

Therefore, \( f'(x) = 2ax - \frac{b}{x^2} \)
In simple words: Expand both f(x + h) and f(x), find their difference, divide by h, factor out common h terms, and then take the limit as h goes to zero.

Exam Tip: When dealing with fractions in the first principle, always find a common denominator before simplifying.

Question 3. (Exercise 28B) Find the derivative of each of the following from the first principle: \( 3x^2 + 2x - 5 \)
Answer: Let \( f(x) = 3x^2 + 2x - 5 \)

We need to find the derivative of f(x), that is, \( f'(x) \)

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots(i) \]

\( f(x) = 3x^2 + 2x - 5 \)
\[ f(x + h) = 3(x + h)^2 + 2(x + h) - 5 = 3(x^2 + h^2 + 2xh) + 2x + 2h - 5 \]
\[ = 3x^2 + 3h^2 + 6xh + 2x + 2h - 5 \]

Putting values in (i), we get:
\[ f'(x) = \lim_{h \to 0} \frac{(3x^2 + 3h^2 + 6xh + 2x + 2h - 5) - (3x^2 + 2x - 5)}{h} \]
\[ = \lim_{h \to 0} \frac{3h^2 + 6xh + 2h}{h} = \lim_{h \to 0} (3h + 6x + 2) \]

Putting h = 0, we get:
\[ f'(x) = 3(0) + 6x + 2 = 6x + 2 \]

Therefore, \( f'(x) = 6x + 2 \)
In simple words: Expand f(x + h) using the algebraic identity for (x + h) squared, subtract f(x), factor out h, and then evaluate the limit.

Exam Tip: Always use the binomial expansion formula correctly: \( (x + h)^2 = x^2 + 2xh + h^2 \). Careless expansion leads to errors.

Question 4. (Exercise 28B) Find the derivative of each of the following from the first principle: \( x^3 - 2x^2 + x + 3 \)
Answer: Let \( f(x) = x^3 - 2x^2 + x + 3 \)

We need to find the derivative of f(x), that is, \( f'(x) \)

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots(i) \]

\( f(x) = x^3 - 2x^2 + x + 3 \)
\[ f(x + h) = (x + h)^3 - 2(x + h)^2 + (x + h) + 3 \]

Putting values in (i), we get:
\[ f'(x) = \lim_{h \to 0} \frac{[(x + h)^3 - 2(x + h)^2 + (x + h) + 3] - [x^3 - 2x^2 + x + 3]}{h} \]
\[ = \lim_{h \to 0} \frac{[(x + h)^3 - x^3] - 2[(x + h)^2 - x^2] + [x + h - x]}{h} \]

Using the identities:
\[ (a + b)^3 = a^3 + b^3 + 3ab^2 + 3a^2b \]
\[ (a + b)^2 = a^2 + b^2 + 2ab \]

We get:
\[ f'(x) = \lim_{h \to 0} \frac{[x^3 + h^3 + 3xh^2 + 3x^2h - x^3] - 2[x^2 + h^2 + 2xh - x^2] + h}{h} \]
\[ = \lim_{h \to 0} \frac{h^3 + 3xh^2 + 3x^2h - 2h^2 - 4xh + h}{h} \]
\[ = \lim_{h \to 0} (h^2 + 3xh + 3x^2 - 2h - 4x + 1) \]

Putting h = 0, we get:
\[ f'(x) = 0 + 0 + 3x^2 - 0 - 4x + 1 = 3x^2 - 4x + 1 \]

Therefore, \( f'(x) = 3x^2 - 4x + 1 \)
In simple words: Apply the binomial expansion formulas for cubes and squares, subtract the original function, divide by h, cancel the h factor, and take the limit.

Exam Tip: Memorising the expansions of \( (a+b)^2 \) and \( (a+b)^3 \) is crucial for solving these first principle problems quickly.

Question 5. (Exercise 28B) Find the derivative of each of the following from the first principle: \( x^8 \)
Answer: Let \( f(x) = x^8 \)

We need to find the derivative of f(x), that is, \( f'(x) \)

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots(i) \]

\( f(x) = x^8 \)
\[ f(x + h) = (x + h)^8 \]

Putting values in (i), we get:
\[ f'(x) = \lim_{h \to 0} \frac{(x + h)^8 - x^8}{h} \]

Using the binomial theorem:
\[ (x + h)^8 = x^8 + 8x^7h + 28x^6h^2 + 56x^5h^3 + \ldots + h^8 \]

\[ f'(x) = \lim_{h \to 0} \frac{8x^7h + 28x^6h^2 + 56x^5h^3 + \ldots + h^8}{h} \]
\[ = \lim_{h \to 0} (8x^7 + 28x^6h + 56x^5h^2 + \ldots + h^7) \]

Putting h = 0, we get:
\[ f'(x) = 8x^7 \]

Therefore, \( f'(x) = 8x^7 \)
In simple words: Use the binomial theorem to expand (x + h) to the eighth power. The x to the eighth terms cancel, divide by h to eliminate one h from each remaining term, and then take the limit as h approaches zero.

Exam Tip: For higher powers like \( x^n \), the derivative is always \( nx^{n-1} \) by the power rule - the first principle derivation confirms this pattern.

Question 6. Find the derivation of each of the following from the first principle:
\( \frac{1}{x^3} \)
Answer: Let \( f(x) = \frac{1}{x^3} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{x^3} \)

\( f(x + h) = \frac{1}{(x + h)^3} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x + h)^3} - \frac{1}{x^3}}{h} \)

\( = \lim_{h \to 0} \frac{(x + h)^{-3} - x^{-3}}{(x + h) - x} \)

[Add and subtract x in denominator]

\( = \lim_{z \to x} \frac{z^{-3} - x^{-3}}{z - x} \) where z = x + h and z → x as h → 0

\( = (-3)x^{-3-1} \left[ \because \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \right] \)

\( = -3x^{-4} \)

\( = - \frac{3}{x^4} \)

Hence, \( f'(x) = -\frac{3}{x^4} \)
In simple words: To find the derivative using first principles, set up the limit definition, simplify the complex fraction, and apply the standard limit formula. The result shows the rate at which the reciprocal cubic function changes.

Exam Tip: Always apply the first principle formula carefully; the key step is recognizing when to use the standard limit rule \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \).

Question 7. Find the derivation of each of the following from the first principle:
\( \frac{1}{x^5} \)
Answer: Let \( f(x) = \frac{1}{x^5} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{x^5} \)

\( f(x + h) = \frac{1}{(x + h)^5} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x + h)^5} - \frac{1}{x^5}}{h} \)

\( = \lim_{h \to 0} \frac{(x + h)^{-5} - x^{-5}}{(x + h) - x} \)

[Add and subtract x in denominator]

\( = \lim_{z \to x} \frac{z^{-5} - x^{-5}}{z - x} \) where z = x + h and z → x as h → 0

\( = (-5)x^{-5-1} \left[ \because \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \right] \)

\( = -5x^{-6} \)

\( = - \frac{5}{x^6} \)

Hence, \( f'(x) = -\frac{5}{x^6} \)
In simple words: Follow the first principles approach by setting up the limit of the difference quotient, simplify using algebraic manipulation, and then employ the fundamental limit rule to arrive at the final derivative expression.

Exam Tip: Watch the exponent carefully when applying the limit rule - the power decreases by 1, and the original power becomes the coefficient with a negative sign.

Question 8. Find the derivation of each of the following from the first principle:
\( \sqrt{ax + b} \)
Answer: Let \( f(x) = \sqrt{ax + b} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \sqrt{ax + b} \)

\( f(x + h) = \sqrt{a(x + h) + b} = \sqrt{ax + ah + b} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\sqrt{ax + ah + b} - \sqrt{ax + b}}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{ax + ah + b} - \sqrt{ax + b} \)

\( = \lim_{h \to 0} \frac{\sqrt{ax + ah + b} - \sqrt{ax + b}}{h} \times \frac{\sqrt{ax + ah + b} + \sqrt{ax + b}}{\sqrt{ax + ah + b} + \sqrt{ax + b}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{ax + ah + b})^2 - (\sqrt{ax + b})^2}{h(\sqrt{ax + ah + b} + \sqrt{ax + b})} \)

\( = \lim_{h \to 0} \frac{ax + ah + b - ax - b}{h(\sqrt{ax + ah + b} + \sqrt{ax + b})} \)

\( = \lim_{h \to 0} \frac{ah}{h(\sqrt{ax + ah + b} + \sqrt{ax + b})} \)

\( = \lim_{h \to 0} \frac{a}{\sqrt{ax + ah + b} + \sqrt{ax + b}} \)

Substituting h = 0, we get

\( = \frac{a}{\sqrt{ax + a(0) + b} + \sqrt{ax + b}} = \frac{a}{\sqrt{ax + b} + \sqrt{ax + b}} = \frac{a}{2\sqrt{ax + b}} \)

Hence, \( f'(x) = \frac{a}{2\sqrt{ax + b}} \)
In simple words: When working with square root functions, rationalize the numerator by using the conjugate. This turns the difference of square roots into a difference of the expressions under the radicals, which simplifies nicely.

Exam Tip: Rationalization is essential for radical expressions - multiplying by the conjugate eliminates the radicals from the numerator and makes the limit straightforward to evaluate.

Question 9. Find the derivation of each of the following from the first principle:
\( \sqrt{5x - 4} \)
Answer: Let \( f(x) = \sqrt{5x - 4} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \sqrt{5x - 4} \)

\( f(x + h) = \sqrt{5(x + h) - 4} = \sqrt{5x + 5h - 4} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\sqrt{5x + 5h - 4} - \sqrt{5x - 4}}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{5x + 5h - 4} - \sqrt{5x - 4} \)

\( = \lim_{h \to 0} \frac{\sqrt{5x + 5h - 4} - \sqrt{5x - 4}}{h} \times \frac{\sqrt{5x + 5h - 4} + \sqrt{5x - 4}}{\sqrt{5x + 5h - 4} + \sqrt{5x - 4}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{5x + 5h - 4})^2 - (\sqrt{5x - 4})^2}{h(\sqrt{5x + 5h - 4} + \sqrt{5x - 4})} \)

\( = \lim_{h \to 0} \frac{5x + 5h - 4 - 5x + 4}{h(\sqrt{5x + 5h - 4} + \sqrt{5x - 4})} \)

\( = \lim_{h \to 0} \frac{5h}{h(\sqrt{5x + 5h - 4} + \sqrt{5x - 4})} \)

\( = \lim_{h \to 0} \frac{5}{\sqrt{5x + 5h - 4} + \sqrt{5x - 4}} \)

Substituting h = 0, we get

\( = \frac{5}{\sqrt{5x + 5(0) - 4} + \sqrt{5x - 4}} = \frac{5}{\sqrt{5x - 4} + \sqrt{5x - 4}} = \frac{5}{2\sqrt{5x - 4}} \)

Hence, \( f'(x) = \frac{5}{2\sqrt{5x - 4}} \)
In simple words: The coefficient inside the square root (which is 5) becomes the numerator, and the denominator is twice the original square root function. This pattern holds for all linear expressions under a square root.

Exam Tip: For radicals with a linear expression inside, the derivative always has the form: (coefficient of x) divided by (2 times the original function).

Question 10. Find the derivation of each of the following from the first principle:
\( \frac{1}{\sqrt{x + 2}} \)
Answer: Let \( f(x) = \frac{1}{\sqrt{x + 2}} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{\sqrt{x + 2}} \)

\( f(x + h) = \frac{1}{\sqrt{x + h + 2}} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h + 2}} - \frac{1}{\sqrt{x + 2}}}{h} \)

\( = \lim_{h \to 0} \frac{\sqrt{x + 2} - \sqrt{x + h + 2}}{(\sqrt{x + h + 2})(\sqrt{x + 2})} \times \frac{1}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{x + 2} - \sqrt{x + h + 2} \)

\( = \lim_{h \to 0} \frac{\sqrt{x + 2} - \sqrt{x + h + 2}}{h(\sqrt{x + h + 2})(\sqrt{x + 2})} \times \frac{\sqrt{x + 2} + \sqrt{x + h + 2}}{\sqrt{x + 2} + \sqrt{x + h + 2}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{x + 2})^2 - (\sqrt{x + h + 2})^2}{h(\sqrt{x + h + 2})(\sqrt{x + 2})(\sqrt{x + 2} + \sqrt{x + h + 2})} \)

\( = \lim_{h \to 0} \frac{x + 2 - x - h - 2}{h(\sqrt{x + h + 2})(\sqrt{x + 2})(\sqrt{x + 2} + \sqrt{x + h + 2})} \)

\( = \lim_{h \to 0} \frac{-h}{h(\sqrt{x + h + 2})(\sqrt{x + 2})(\sqrt{x + 2} + \sqrt{x + h + 2})} \)

\( = \lim_{h \to 0} \frac{-1}{(\sqrt{x + h + 2})(\sqrt{x + 2})(\sqrt{x + 2} + \sqrt{x + h + 2})} \)

Substituting h = 0, we get

\( = \frac{-1}{(\sqrt{x + 0 + 2})(\sqrt{x + 2})(\sqrt{x + 2} + \sqrt{x + 0 + 2})} = \frac{-1}{(\sqrt{x + 2})^2(2\sqrt{x + 2})} = \frac{-1}{2(\sqrt{x + 2})^3} \)

Hence, \( f'(x) = \frac{-1}{2(\sqrt{x + 2})^3} \)
In simple words: For reciprocal square root functions, rationalize and simplify carefully. The result is the negative of the reciprocal function raised to a higher power, divided by twice the power.

Exam Tip: When handling reciprocal radical functions, pay attention to the sign - it will be negative. Always verify by checking dimensions and limiting behavior.

Question 11. Find the derivation of each of the following from the first principle:
\( \frac{1}{\sqrt{2x + 3}} \)
Answer: Let \( f(x) = \frac{1}{\sqrt{2x + 3}} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{\sqrt{2x + 3}} \)

\( f(x + h) = \frac{1}{\sqrt{2x + 2h + 3}} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2x + 2h + 3}} - \frac{1}{\sqrt{2x + 3}}}{h} \)

\( = \lim_{h \to 0} \frac{\sqrt{2x + 3} - \sqrt{2x + 2h + 3}}{(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})} \times \frac{1}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{2x + 3} - \sqrt{2x + 2h + 3} \)

\( = \lim_{h \to 0} \frac{\sqrt{2x + 3} - \sqrt{2x + 2h + 3}}{h(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})} \times \frac{\sqrt{2x + 3} + \sqrt{2x + 2h + 3}}{\sqrt{2x + 3} + \sqrt{2x + 2h + 3}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{2x + 3})^2 - (\sqrt{2x + 2h + 3})^2}{h(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \)

\( = \lim_{h \to 0} \frac{2x + 3 - 2x - 2h - 3}{h(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \)

\( = \lim_{h \to 0} \frac{-2h}{h(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \)

\( = \lim_{h \to 0} \frac{-2}{(\sqrt{2x + 2h + 3})(\sqrt{2x + 3})(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})} \)

Substituting h = 0, we get

\( = \frac{-2}{(\sqrt{2x + 0 + 3})(\sqrt{2x + 3})(\sqrt{2x + 3} + \sqrt{2x + 0 + 3})} = \frac{-2}{(\sqrt{2x + 3})^2(2\sqrt{2x + 3})} = \frac{-2}{2(\sqrt{2x + 3})^3} = \frac{-1}{(\sqrt{2x + 3})^3} \)

Hence, \( f'(x) = \frac{-1}{(\sqrt{2x + 3})^3} \)
In simple words: The coefficient of x inside the radical (which is 2) becomes part of the numerator during simplification, but ultimately cancels to give a cleaner result with the base expression raised to the third power in the denominator.

Exam Tip: Always track the coefficient inside the radical carefully - it will appear in the numerator after rationalization and help determine the final simplified form.

Question 12. Find the derivation of each of the following from the first principle:
\( \frac{1}{\sqrt{6x - 5}} \)
Answer: Let \( f(x) = \frac{1}{\sqrt{6x - 5}} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{\sqrt{6x - 5}} \)

\( f(x + h) = \frac{1}{\sqrt{6x + 6h - 5}} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{6x + 6h - 5}} - \frac{1}{\sqrt{6x - 5}}}{h} \)

\( = \lim_{h \to 0} \frac{\sqrt{6x - 5} - \sqrt{6x + 6h - 5}}{(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})} \times \frac{1}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{6x - 5} - \sqrt{6x + 6h - 5} \)

\( = \lim_{h \to 0} \frac{\sqrt{6x - 5} - \sqrt{6x + 6h - 5}}{h(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})} \times \frac{\sqrt{6x - 5} + \sqrt{6x + 6h - 5}}{\sqrt{6x - 5} + \sqrt{6x + 6h - 5}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{6x - 5})^2 - (\sqrt{6x + 6h - 5})^2}{h(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})(\sqrt{6x - 5} + \sqrt{6x + 6h - 5})} \)

\( = \lim_{h \to 0} \frac{6x - 5 - 6x - 6h + 5}{h(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})(\sqrt{6x - 5} + \sqrt{6x + 6h - 5})} \)

\( = \lim_{h \to 0} \frac{-6h}{h(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})(\sqrt{6x - 5} + \sqrt{6x + 6h - 5})} \)

\( = \lim_{h \to 0} \frac{-6}{(\sqrt{6x + 6h - 5})(\sqrt{6x - 5})(\sqrt{6x - 5} + \sqrt{6x + 6h - 5})} \)

Substituting h = 0, we get

\( = \frac{-6}{(\sqrt{6x + 6(0) - 5})(\sqrt{6x - 5})(\sqrt{6x - 5} + \sqrt{6x + 6(0) - 5})} = \frac{-6}{(\sqrt{6x - 5})^2(2\sqrt{6x - 5})} = \frac{-6}{2(\sqrt{6x - 5})^3} = \frac{-3}{(\sqrt{6x - 5})^3} \)

Hence, \( f'(x) = \frac{-3}{(\sqrt{6x - 5})^3} \)
In simple words: The coefficient in front of x (which is 6) divides by 2 to yield the final coefficient of 3 in the numerator, producing a negative result with the radical expression cubed in the denominator.

Exam Tip: When the coefficient of x is even, dividing it by 2 gives the final coefficient in the answer - this is a useful shortcut to check your work mentally.

Question 13. Find the derivation of each of the following from the first principle:
\( \frac{1}{\sqrt{2 - 3x}} \)
Answer: Let \( f(x) = \frac{1}{\sqrt{2 - 3x}} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{1}{\sqrt{2 - 3x}} \)

\( f(x + h) = \frac{1}{\sqrt{2 - 3(x + h)}} = \frac{1}{\sqrt{2 - 3x - 3h}} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2 - 3x - 3h}} - \frac{1}{\sqrt{2 - 3x}}}{h} \)

\( = \lim_{h \to 0} \frac{\sqrt{2 - 3x} - \sqrt{2 - 3x - 3h}}{(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})} \times \frac{1}{h} \)

Rationalize the numerator by multiplying and dividing by the conjugate of \( \sqrt{2 - 3x} - \sqrt{2 - 3x - 3h} \)

\( = \lim_{h \to 0} \frac{\sqrt{2 - 3x} - \sqrt{2 - 3x - 3h}}{h(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})} \times \frac{\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h}}{\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h}} \)

Using the formula: (a + b)(a - b) = (a² - b²)

\( = \lim_{h \to 0} \frac{(\sqrt{2 - 3x})^2 - (\sqrt{2 - 3x - 3h})^2}{h(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})(\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h})} \)

\( = \lim_{h \to 0} \frac{2 - 3x - 2 + 3x + 3h}{h(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})(\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h})} \)

\( = \lim_{h \to 0} \frac{3h}{h(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})(\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h})} \)

\( = \lim_{h \to 0} \frac{3}{(\sqrt{2 - 3x - 3h})(\sqrt{2 - 3x})(\sqrt{2 - 3x} + \sqrt{2 - 3x - 3h})} \)

Substituting h = 0, we get

\( = \frac{3}{(\sqrt{2 - 3x - 3(0)})(\sqrt{2 - 3x})(\sqrt{2 - 3x} + \sqrt{2 - 3x - 3(0)})} = \frac{3}{(\sqrt{2 - 3x})^2(2\sqrt{2 - 3x})} = \frac{3}{2(\sqrt{2 - 3x})^3} \)

Hence, \( f'(x) = \frac{3}{2(\sqrt{2 - 3x})^3} \)
In simple words: When the coefficient of x is negative, the subtraction in the numerator reverses, making the result positive. The magnitude of that coefficient (3) appears in the final answer.

Exam Tip: Always be careful with negative coefficients - track the signs through rationalization to ensure your final answer has the correct sign.

Question 14. Find the derivation of each of the following from the first principle:
\( \frac{2x + 3}{3x + 2} \)
Answer: Let \( f(x) = \frac{2x + 3}{3x + 2} \). We need to determine the derivative of f(x), which is f'(x).

We understand that

\( f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) …(i)

\( f(x) = \frac{2x + 3}{3x + 2} \)

\( f(x + h) = \frac{2(x + h) + 3}{3(x + h) + 2} = \frac{2x + 2h + 3}{3x + 3h + 2} \)

Substituting these into (i), we get

\( f'(x) = \lim_{h \to 0} \frac{\frac{2x + 2h + 3}{3x + 3h + 2} - \frac{2x + 3}{3x + 2}}{h} \)

Combine the fractions in the numerator:

\( = \lim_{h \to 0} \frac{(2x + 2h + 3)(3x + 2) - (2x + 3)(3x + 3h + 2)}{h(3x + 3h + 2)(3x + 2)} \)

Expand the numerator:

(2x + 2h + 3)(3x + 2) = 6x² + 4x + 6hx + 4h + 9x + 6

(2x + 3)(3x + 3h + 2) = 6x² + 6hx + 4x + 9x + 9h + 6

Subtract:

\( = \lim_{h \to 0} \frac{6x^2 + 4x + 6hx + 4h + 9x + 6 - 6x^2 - 6hx - 4x - 9x - 9h - 6}{h(3x + 3h + 2)(3x + 2)} \)

\( = \lim_{h \to 0} \frac{4h - 9h}{h(3x + 3h + 2)(3x + 2)} \)

\( = \lim_{h \to 0} \frac{-5h}{h(3x + 3h + 2)(3x + 2)} \)

\( = \lim_{h \to 0} \frac{-5}{(3x + 3h + 2)(3x + 2)} \)

Substituting h = 0, we get

\( = \frac{-5}{(3x + 3(0) + 2)(3x + 2)} = \frac{-5}{(3x + 2)^2} \)

Hence, \( f'(x) = \frac{-5}{(3x + 2)^2} \)
In simple words: For rational functions, find a common denominator, expand both products in the numerator, subtract carefully to isolate the h term, then cancel the h and evaluate the limit at h = 0.

Exam Tip: When expanding products of binomials, use FOIL or distribution carefully - small errors in arithmetic can lead to missing the h term that allows cancellation.

Question 15. Find the derivation of each of the following from the first principle:
\( \frac{5-x}{5+x} \)
Answer: Let \( f(x) = \frac{5-x}{5+x} \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \frac{5-x}{5+x} \)

\( f(x + h) = \frac{5 - (x + h)}{5 + (x + h)} = \frac{5 - x - h}{5 + x + h} \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\frac{5-x-h}{5+x+h} - \frac{5-x}{5+x}}{h} \]

\[ = \lim_{h \to 0} \frac{(5-x-h)(5+x) - (5-x)(5+x+h)}{(5+x+h)(5+x) \cdot h} \]

\[ = \lim_{h \to 0} \frac{25 + 5x - 5x - x^2 - 5h - xh - [25 + 5h + 5x - x^2 - xh]}{h(5+x+h)(5+x)} \]

\[ = \lim_{h \to 0} \frac{25 - x^2 - 5h - xh - 25 - 5h + x^2 + xh}{h(5+x+h)(5+x)} \]

\[ = \lim_{h \to 0} \frac{-10h}{h(5+x+h)(5+x)} \]

\[ = \lim_{h \to 0} \frac{-10}{(5+x+h)(5+x)} \]

Setting h = 0,
\[ = \frac{-10}{(5+x+0)(5+x)} = \frac{-10}{(5+x)(5+x)} = \frac{-10}{(5+x)^2} \]

Hence,
\[ f'(x) = \frac{-10}{(5+x)^2} \]
In simple words: Apply the first principle formula to calculate the derivative. After simplifying using algebra and taking the limit as h approaches zero, you arrive at the final answer.

Exam Tip: Always simplify the difference quotient fully before taking the limit - factoring out h from the numerator is crucial for avoiding the indeterminate 0/0 form.

Question 16. Find the derivation of each of the following from the first principle:
\( \frac{x^2+1}{x}, x \neq 0 \)
Answer: Let \( f(x) = \frac{x^2+1}{x} \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \frac{x^2+1}{x} \)

\( f(x+h) = \frac{(x+h)^2+1}{x+h} = \frac{x^2 + h^2 + 2xh + 1}{x+h} \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\frac{x^2+h^2+2xh+1}{x+h} - \frac{x^2+1}{x}}{h} \]

\[ = \lim_{h \to 0} \frac{(x^2 + h^2 + 2xh + 1)(x) - (x^2 + 1)(x + h)}{(x + h)(x) \cdot h} \]

\[ = \lim_{h \to 0} \frac{x^3 + xh^2 + 2x^2h + x - [x^3 + x^2h + x + h]}{h(x+h)(x)} \]

\[ = \lim_{h \to 0} \frac{x^3 + xh^2 + 2x^2h + x - x^3 - x^2h - x - h}{h(x+h)(x)} \]

\[ = \lim_{h \to 0} \frac{xh^2 + x^2h - h}{h(x+h)(x)} \]

\[ = \lim_{h \to 0} \frac{xh + x^2 - 1}{(x+h)(x)} \]

Setting h = 0,
\[ = \frac{x(0) + x^2 - 1}{(x + 0)(x)} = \frac{x^2 - 1}{x^2} \]

Hence,
\[ f'(x) = \frac{x^2 - 1}{x^2} \]
In simple words: Use the first principle to compute the derivative. Simplify by combining fractions, then divide through by h and take the limit.

Exam Tip: When dealing with rational functions, find a common denominator before simplifying - this prevents algebraic errors during cancellation.

Question 17. Find the derivation of each of the following from the first principle:
\( \sqrt{\cos 3x} \)
Answer: Let \( f(x) = \sqrt{\cos 3x} \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \sqrt{\cos 3x} \)

\( f(x+h) = \sqrt{\cos 3(x+h)} = \sqrt{\cos(3x + 3h)} \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\cos(3x + 3h)} - \sqrt{\cos 3x}}{h} \]

Now rationalizing the numerator by multiplying and dividing by the conjugate of \( \sqrt{\cos(3x + 3h)} - \sqrt{\cos 3x} \)

\[ = \lim_{h \to 0} \frac{\sqrt{\cos(3x + 3h)} - \sqrt{\cos 3x}}{h} \times \frac{\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x}}{\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x}} \]

Using the formula: \( (a + b)(a - b) = (a^2 - b^2) \)

\[ = \lim_{h \to 0} \frac{(\sqrt{\cos(3x + 3h)})^2 - (\sqrt{\cos 3x})^2}{h(\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x})} \]

\[ = \lim_{h \to 0} \frac{\cos(3x + 3h) - \cos 3x}{h(\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x})} \]

Using the formula: \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)

\[ = \lim_{h \to 0} \frac{-2 \sin\left(\frac{3x + 3h + 3x}{2}\right) \sin\left(\frac{3x + 3h - 3x}{2}\right)}{h(\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x})} \]

\[ = \lim_{h \to 0} \frac{-2 \sin\left(\frac{6x + 3h}{2}\right) \sin\left(\frac{3h}{2}\right)}{h(\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x})} \]

\[ = -2 \lim_{h \to 0} \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \times \frac{3}{2} \lim_{h \to 0} \sin\left(\frac{6x + 3h}{2}\right) \times \lim_{h \to 0} \frac{1}{\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x}} \]

[Here, we multiply and divide by \( \frac{3}{2} \)]

\[ = -3 \times (1) \times \lim_{h \to 0} \sin\left(\frac{6x + 3h}{2}\right) \times \lim_{h \to 0} \frac{1}{\sqrt{\cos(3x + 3h)} + \sqrt{\cos 3x}} \]

\[ \left[\because \lim_{x \to 0} \frac{\sin x}{x} = 1\right] \]

Setting h = 0,
\[ = -3 \times \sin\left(\frac{6x + 3(0)}{2}\right) \times \frac{1}{\sqrt{\cos(3x + 3(0))} + \sqrt{\cos 3x}} \]

\[ = -3 \sin 3x \times \frac{1}{2\sqrt{\cos 3x}} \]

\[ = -\frac{3\sin 3x}{2\sqrt{\cos 3x}} = -\frac{3\sin 3x}{2(\cos 3x)^{1/2}} \]

Hence,
\[ f'(x) = -\frac{3\sin 3x}{2(\cos 3x)^{1/2}} \]
In simple words: Rationalize the difference between square roots, then apply trigonometric identities to simplify. Use the standard limit \( \lim \frac{\sin x}{x} = 1 \) before substituting h = 0.

Exam Tip: When rationalizing square root expressions in the first principle, multiply by the conjugate immediately - this converts the difference of square roots into a simpler algebraic form.

Question 18. Find the derivation of each of the following from the first principle:
\( \sqrt{\sec x} \)
Answer: Let \( f(x) = \sqrt{\sec x} \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \sqrt{\sec x} \)

\( f(x+h) = \sqrt{\sec(x+h)} \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\sec(x+h)} - \sqrt{\sec x}}{h} \]

Now rationalizing the numerator by multiplying and dividing by the conjugate of \( \sqrt{\sec(x+h)} - \sqrt{\sec x} \)

\[ = \lim_{h \to 0} \frac{\sqrt{\sec(x+h)} - \sqrt{\sec x}}{h} \times \frac{\sqrt{\sec(x+h)} + \sqrt{\sec x}}{\sqrt{\sec(x+h)} + \sqrt{\sec x}} \]

Using the formula: \( (a + b)(a - b) = (a^2 - b^2) \)

\[ = \lim_{h \to 0} \frac{(\sqrt{\sec(x+h)})^2 - (\sqrt{\sec x})^2}{h(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = \lim_{h \to 0} \frac{\sec(x+h) - \sec(x)}{h(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = \lim_{h \to 0} \frac{\frac{1}{\cos(x+h)} - \frac{1}{\cos x}}{h(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = \lim_{h \to 0} \frac{\frac{\cos x - \cos(x+h)}{\cos(x+h) \cos x}}{h(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = \lim_{h \to 0} \frac{\cos x - \cos(x+h)}{h(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

Using the formula: \( \cos A - \cos B = 2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{B-A}{2}\right) \)

\[ = \lim_{h \to 0} \frac{2 \sin\left(\frac{x + (x+h)}{2}\right) \sin\left(\frac{(x+h) - x}{2}\right)}{h(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = \lim_{h \to 0} \frac{2 \sin\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = 2\lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \times \frac{1}{2} \lim_{h \to 0} \sin\left(\frac{2x + h}{2}\right) \times \lim_{h \to 0} \frac{1}{(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

[Here, we multiply and divide by \( \frac{1}{2} \)]

\[ = 2 \times \frac{1}{2} \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \times \frac{1}{2} \lim_{h \to 0} \sin\left(\frac{2x + h}{2}\right) \times \lim_{h \to 0} \frac{1}{(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ = (1) \times \lim_{h \to 0} \sin\left(x + \frac{h}{2}\right) \times \lim_{h \to 0} \frac{1}{(\cos(x+h) \cos x)(\sqrt{\sec(x+h)} + \sqrt{\sec x})} \]

\[ \left[\because \lim_{x \to 0} \frac{\sin x}{x} = 1\right] \]

Setting h = 0,
\[ = \sin\left[x + \frac{0}{2}\right] \times \frac{1}{(\cos(x + 0) \cos x)(\sqrt{\sec(x + 0)} + \sqrt{\sec x})} \]

\[ = \sin x \times \frac{1}{\cos x \cos x(\sqrt{\sec x} + \sqrt{\sec x})} \]

\[ = \frac{\sin x}{\cos^2 x(2\sqrt{\sec x})} \]

\[ = \frac{\sin x}{\cos x} \times \frac{1}{\cos x} \times \frac{1}{2\sqrt{\sec x}} \]

\[ = \tan x \times \sec x \times \frac{1}{2\sqrt{\sec x}} \quad \left[\because \frac{\sin x}{\cos x} = \tan x \text{ and } \frac{1}{\cos x} = \sec x\right] \]

\[ = \frac{1}{2} \tan x \sqrt{\sec x} \]

Hence,
\[ f'(x) = \frac{1}{2} \tan x \sqrt{\sec x} \]
In simple words: Rationalize by multiplying by the conjugate, convert secant to cosine, use the cosine difference formula, and apply the standard limit before substituting h = 0.

Exam Tip: Break down the final expression carefully using trigonometric identities - verify each identity substitution before proceeding to ensure no errors accumulate.

Question 19. Find the derivation of each of the following from the first principle:
\( \tan^2 x \)
Answer: Let \( f(x) = \tan^2 x \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \tan^2 x \)

\( f(x+h) = \tan^2(x+h) \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\tan^2(x+h) - \tan^2 x}{h} \]

\[ = \lim_{h \to 0} \frac{[\tan(x+h) - \tan x][\tan(x+h) + \tan x]}{h} \]

Using: \( \tan x = \frac{\sin x}{\cos x} \)

\[ = \lim_{h \to 0} \frac{\left[\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}\right]\left[\frac{\sin(x+h)}{\cos(x+h)} + \frac{\sin x}{\cos x}\right]}{h} \]

\[ = \lim_{h \to 0} \frac{\left[\frac{\sin(x+h)\cos x - \sin x \cos(x+h)}{\cos(x+h) \cos x}\right]\left[\frac{\sin(x+h)\cos x + \sin x \cos(x+h)}{\cos(x+h) \cos x}\right]}{h} \]

\[ = \lim_{h \to 0} \frac{\{\sin[(x+h) - x]\}\{\sin[(x+h) + x]\}}{h[\cos^2(x+h) \cos^2 x]} \]

\[ \left[\because \sin A \cos B - \sin B \cos A = \sin(A - B) \text{ and } \sin A \cos B + \sin B \cos A = \sin(A + B)\right] \]

\[ = \lim_{h \to 0} \frac{\sin h \sin(2x+h)}{h[\cos^2(x+h) \cos^2 x]} \]

\[ = \frac{1}{\cos^2 x} \lim_{h \to 0} \frac{\sin h}{h} \times \lim_{h \to 0} \sin(2x+h) \times \lim_{h \to 0} \frac{1}{\cos^2(x+h)} \]

\[ = \frac{1}{\cos^2 x} \times (1) \times \lim_{h \to 0} \sin(2x+h) \times \lim_{h \to 0} \frac{1}{\cos^2(x+h)} \]

\[ \left[\because \lim_{x \to 0} \frac{\sin x}{x} = 1\right] \]

Setting h = 0,
\[ = \frac{1}{\cos^2 x} \times \sin(2x + 0) \times \frac{1}{\cos^2(x + 0)} \]

\[ = \frac{1}{\cos^2 x} \times \sin 2x \times \frac{1}{\cos^2 x} \]

\[ = \frac{1}{\cos^2 x} \times 2\sin x \cos x \times \sec^2 x \quad [\because \sin 2x = 2\sin x \cos x] \]

\[ = 2 \frac{\sin x}{\cos x} \times \sec^2 x \quad \left[\because \frac{\sin x}{\cos x} = \tan x\right] \]

\[ = 2\tan x \sec^2 x \]

Hence, \( f'(x) = 2\tan x \sec^2 x \)
In simple words: Factor the difference of squares, express tangent in terms of sine and cosine, apply trigonometric identities for sum and difference of angles, then take the limit after canceling h.

Exam Tip: Always factor \( a^2 - b^2 \) as \( (a-b)(a+b) \) first - this separation is key for isolating the h-dependent term that allows cancellation.

Question 20. Find the derivation of each of the following from the first principle:
\( \sin(2x + 3) \)
Answer: Let \( f(x) = \sin(2x + 3) \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \sin(2x + 3) \)

\( f(x+h) = \sin[2(x+h) + 3] \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\sin[2(x+h) + 3] - \sin(2x + 3)}{h} \]

Using the formula: \( \sin A - \sin B = 2 \sin\left(\frac{A-B}{2}\right) \cos\left(\frac{A+B}{2}\right) \)

\[ = \lim_{h \to 0} \frac{2 \sin\left(\frac{2(x+h) + 3 - (2x + 3)}{2}\right) \cos\left(\frac{2(x+h) + 3 + (2x + 3)}{2}\right)}{h} \]

\[ = \lim_{h \to 0} \frac{2 \sin\left(\frac{2x + 2h + 3 - 2x - 3}{2}\right) \cos\left(\frac{2x + 2h + 3 + 2x + 3}{2}\right)}{h} \]

\[ = \lim_{h \to 0} \frac{2 \sin\left(\frac{2h}{2}\right) \cos\left(\frac{4x + 2h + 6}{2}\right)}{h} \]

\[ = \lim_{h \to 0} \frac{2 \sin(h) \cos(2x + h + 3)}{h} \]

\[ = 2\lim_{h \to 0} \frac{\sin h}{h} \times \lim_{h \to 0} \cos(2x + h + 3) \]

\[ = 2(1) \times \lim_{h \to 0} \cos(2x + h + 3) \]

\[ \left[\because \lim_{x \to 0} \frac{\sin x}{x} = 1\right] \]

Setting h = 0,
\[ = 2 \times \cos(2x + 0 + 3) \]

\[ = 2\cos(2x + 3) \]

Hence, \( f'(x) = 2\cos(2x + 3) \)
In simple words: Use the sine difference formula, extract the standard limit \( \lim \frac{\sin h}{h} = 1 \), then substitute h = 0 to get the final answer.

Exam Tip: When a linear argument like (2x + 3) appears inside a trigonometric function, the coefficient of x (which is 2 here) always appears in the derivative - verify this matches your final answer.

Question 21. Find the derivation of each of the following from the first principle:
\( \tan(3x + 1) \)
Answer: Let \( f(x) = \tan(3x + 1) \)

We need to find the derivative of f(x), that is, f'(x).

We know that,
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \ldots (i) \]

\( f(x) = \tan(3x + 1) \)

\( f(x+h) = \tan[3(x+h) + 1] \)

Substituting into (i),
\[ f'(x) = \lim_{h \to 0} \frac{\tan[3(x+h) + 1] - \tan(3x + 1)}{h} \]

Using the formula: \( \tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B} \)

\[ = \lim_{h \to 0} \frac{\frac{\sin[3(x+h) + 1 - (3x + 1)]}{\cos[3(x+h) + 1] \cos[3x + 1]}}{h} \]

\[ = \lim_{h \to 0} \frac{\sin[3x + 3h + 1 - 3x - 1]}{h \cos[3(x+h) + 1] \cos[3x + 1]} \]

\[ = \lim_{h \to 0} \frac{\sin 3h}{h \cos[3(x+h) + 1] \cos[3x + 1]} \]

\[ = \lim_{h \to 0} \frac{\sin 3h}{3h} \times 3 \times \lim_{h \to 0} \frac{1}{\cos[3(x+h) + 1] \cos[3x + 1]} \]

\[ = 3(1) \times \lim_{h \to 0} \frac{1}{\cos[3(x+h) + 1] \cos[3x + 1]} \]

\[ \left[\because \lim_{x \to 0} \frac{\sin 3x}{3x} = 1\right] \]

Setting h = 0,
\[ = 3 \times \frac{1}{\cos[3(x + 0) + 1] \cos[3x + 1]} \]

\[ = \frac{3}{\cos[3x + 1] \cos[3x + 1]} \]

\[ = \frac{3}{\cos^2(3x + 1)} \]

\[ = 3\sec^2(3x + 1) \quad \left[\because \frac{1}{\cos x} = \sec x\right] \]

Hence, \( f'(x) = 3\sec^2(3x + 1) \)
In simple words: Apply the tangent difference formula, isolate the \( \lim \frac{\sin 3h}{3h} = 1 \) form, then set h = 0 and express the result using the secant function.

Exam Tip: For composite functions like \( \tan(3x + 1) \), the derivative picks up the coefficient of x (which is 3) as a multiplier - if you get \( \sec^2(3x + 1) \) without the 3, you made an error in extracting the standard limit form.

Exercise 28C

Question 1. Differentiate: x² sin x
Answer: To find the derivative of x² sin x, we apply the product rule.

Formula used: \( (uv)' = u'v + uv' \) (Leibnitz or product rule)

Let u = x² and v = sin x

Then u' = 2x and v' = cos x

Using the product rule,
\[ \frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) \]

\[ = 2x \sin x + x^2 \cos x \]

Hence, \( \frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x \)
In simple words: Take the derivative of the first function, multiply by the second unchanged, then add the first unchanged multiplied by the derivative of the second.

Exam Tip: Always identify which part is u and which is v before applying the product rule - clearly labeling makes it harder to mix up the formula or forget a term.

Question 1. Differentiate \( x^2 \sin x \)
Answer: To find the derivative of \( x^2 \sin x \), we apply the product rule.

Let \( u = x^2 \) and \( v = \sin x \).

Then \( u' = 2x \) and \( v' = \cos x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (x^2 \sin x)' = 2x \times \sin x + x^2 \times \cos x = 2x \sin x + x^2 \cos x \)

In simple words: Multiply the derivative of the first part by the second part, then add the first part multiplied by the derivative of the second part.

Exam Tip: Always remember the product rule structure: first derivative times second function, plus first function times second derivative. Check your final answer by factoring if possible.

Question 2. Differentiate \( e^x \cos x \)
Answer: To find the derivative of \( e^x \cos x \), we apply the product rule.

Let \( u = e^x \) and \( v = \cos x \).

Then \( u' = e^x \) and \( v' = -\sin x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (e^x \cos x)' = e^x \times \cos x + e^x \times (-\sin x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x) \)

In simple words: Since \( e^x \) appears in both terms, we can factor it out to simplify the final result.

Exam Tip: Watch for the negative sign when differentiating cosine. Always look for opportunities to factor common terms in your final answer.

Question 3. Differentiate \( e^x \cot x \)
Answer: To find the derivative of \( e^x \cot x \), we apply the product rule.

Let \( u = e^x \) and \( v = \cot x \).

Then \( u' = e^x \) and \( v' = -\csc^2 x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (e^x \cot x)' = e^x \times \cot x + e^x \times (-\csc^2 x) = e^x \cot x - e^x \csc^2 x = e^x(\cot x - \csc^2 x) \)

In simple words: The exponential function \( e^x \) is retained in each term, allowing us to factor it from both parts.

Exam Tip: Recall that the derivative of \( \cot x \) is \( -\csc^2 x \). Factoring out the common exponential term shows good mathematical practice and simplifies the expression.

Question 4. Differentiate \( x^n \cot x \)
Answer: To find the derivative of \( x^n \cot x \), we apply the product rule.

Let \( u = x^n \) and \( v = \cot x \).

Then \( u' = nx^{n-1} \) and \( v' = -\csc^2 x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (x^n \cot x)' = nx^{n-1} \times \cot x + x^n \times (-\csc^2 x) = nx^{n-1} \cot x - x^n \csc^2 x = x^n(nx^{-1} \cot x - \csc^2 x) \)

In simple words: Apply the power rule to \( x^n \) and use the known derivative of cotangent, then combine using the product rule.

Exam Tip: When factoring \( x^n \) from both terms, remember to adjust the exponent accordingly in the factored form. This technique keeps expressions cleaner.

Question 5. Differentiate \( x^3 \sec x \)
Answer: To find the derivative of \( x^3 \sec x \), we apply the product rule.

Let \( u = x^3 \) and \( v = \sec x \).

Then \( u' = 3x^2 \) and \( v' = \sec x \tan x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (x^3 \sec x)' = 3x^2 \times \sec x + x^3 \times \sec x \tan x = 3x^2 \sec x + x^3 \sec x \tan x = x^2 \sec x(3 + x \tan x) \)

In simple words: Use the power rule on the polynomial and the trigonometric derivative on the secant function, then factor out the common terms.

Exam Tip: The derivative of \( \sec x \) is \( \sec x \tan x \). Factoring \( x^2 \sec x \) from both terms shows the structure of the answer clearly.

Question 6. Differentiate \( (x^2 + 3x + 1) \sin x \)
Answer: To find the derivative of \( (x^2 + 3x + 1) \sin x \), we apply the product rule.

Let \( u = x^2 + 3x + 1 \) and \( v = \sin x \).

Then \( u' = 2x + 3 \) and \( v' = \cos x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( [(x^2 + 3x + 1) \sin x]' = (2x + 3) \times \sin x + (x^2 + 3x + 1) \times \cos x = (2x + 3) \sin x + (x^2 + 3x + 1) \cos x \)

In simple words: Differentiate the polynomial part separately, apply the product rule, and write the answer with both terms clearly shown.

Exam Tip: Be careful when differentiating polynomial expressions with multiple terms. The product rule applies to the entire polynomial as a single unit, not to each term separately.

Question 7. Differentiate \( x^4 \tan x \)
Answer: To find the derivative of \( x^4 \tan x \), we apply the product rule.

Let \( u = x^4 \) and \( v = \tan x \).

Then \( u' = 4x^3 \) and \( v' = \sec^2 x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( (x^4 \tan x)' = 4x^3 \times \tan x + x^4 \times \sec^2 x = 4x^3 \tan x + x^4 \sec^2 x = x^3(4 \tan x + x \sec^2 x) \)

In simple words: Use the power rule for the polynomial and recall that the derivative of tangent is secant squared, then factor out \( x^3 \).

Exam Tip: Always verify the derivative of tangent is \( \sec^2 x \) before working. Factoring out the highest power of \( x \) makes your final answer more compact and professional.

Question 8. Differentiate \( (3x - 5)(4x^2 - 3 + e^x) \)
Answer: To find the derivative of \( (3x - 5)(4x^2 - 3 + e^x) \), we apply the product rule.

Let \( u = 3x - 5 \) and \( v = 4x^2 - 3 + e^x \).

Then \( u' = 3 \) and \( v' = 8x + e^x \).

Using the product rule \( (uv)' = u'v + uv' \):

\( [(3x - 5)(4x^2 - 3 + e^x)]' = 3(4x^2 - 3 + e^x) + (3x - 5)(8x + e^x) \)

\( = 12x^2 - 9 + 3e^x + 24x^2 + 3xe^x - 40x - 5e^x = 36x^2 + x(3e^x - 40) - 9 - 2e^x \)

In simple words: Differentiate the first polynomial to get 3, differentiate the second to get \( 8x + e^x \), apply the product rule, expand carefully, and combine like terms.

Exam Tip: Expand each product fully and group similar terms together at the end. Keep track of all \( e^x \) terms separately since they behave differently than polynomial terms.

Question 9. Differentiate \( (x^2 - 4x + 5)(x^3 - 2) \)
Answer: To find the derivative of \( (x^2 - 4x + 5)(x^3 - 2) \), we apply the product rule.

Let \( u = x^2 - 4x + 5 \) and \( v = x^3 - 2 \).

Then \( u' = 2x - 4 \) and \( v' = 3x^2 \).

Using the product rule \( (uv)' = u'v + uv' \):

\( [(x^2 - 4x + 5)(x^3 - 2)]' = (2x - 4)(x^3 - 2) + (x^2 - 4x + 5)(3x^2) \)

\( = 2x^4 - 4x - 4x^3 + 8 + 3x^4 - 12x^3 + 15x^2 = 5x^4 - 16x^3 + 15x^2 - 4x + 8 \)

In simple words: Differentiate each polynomial separately, multiply accordingly by the product rule, expand both products, then add them together and simplify by combining all similar terms.

Exam Tip: Write out the expansion step by step to avoid sign errors. Arrange your final answer in descending order of powers for clarity and professional presentation.

Question 10. Differentiate \( (x^2 + 2x - 3)(x^2 + 7x + 5) \)
Answer: To find the derivative of \( (x^2 + 2x - 3)(x^2 + 7x + 5) \), we apply the product rule.

Let \( u = x^2 + 2x - 3 \) and \( v = x^2 + 7x + 5 \).

Then \( u' = 2x + 2 \) and \( v' = 2x + 7 \).

Using the product rule \( (uv)' = u'v + uv' \):

\( [(x^2 + 2x - 3)(x^2 + 7x + 5)]' = (2x + 2)(x^2 + 7x + 5) + (x^2 + 2x - 3)(2x + 7) \)

\( = 2x^3 + 14x^2 + 10x + 2x^2 + 14x + 10 + 2x^3 + 7x^2 + 4x^2 + 14x - 6x - 21 = 4x^3 + 27x^2 + 32x - 11 \)

In simple words: Use the product rule by differentiating each quadratic and multiplying with the other factor, then expand both products and collect all terms of the same degree.

Exam Tip: When working with two similar polynomial degrees, be extra careful to combine like terms systematically. Double-check your arithmetic by counting the total number of terms before and after simplification.

Question 11. Differentiate \( (\tan x + \sec x)(\cot x + \csc x) \)
Answer: To find the derivative of \( (\tan x + \sec x)(\cot x + \csc x) \), we apply the product rule.

Let \( u = \tan x + \sec x \) and \( v = \cot x + \csc x \).

Then \( u' = \sec^2 x + \sec x \tan x = \sec x(\sec x + \tan x) \) and \( v' = -\csc^2 x - \csc x \cot x = -\csc x(\csc x + \cot x) \).

Using the product rule \( (uv)' = u'v + uv' \):

\( [(\tan x + \sec x)(\cot x + \csc x)]' = \sec x(\sec x + \tan x)(\cot x + \csc x) - \csc x(\sec x + \tan x)(\csc x + \cot x) \)

\( = (\sec x + \tan x)[\sec x(\cot x + \csc x) - \csc x(\csc x + \cot x)] = (\sec x + \tan x)(\sec x - \csc x)(\cot x + \csc x) \)

In simple words: Factor out the common trigonometric expressions from the resulting terms after applying the product rule to make the final answer cleaner.

Exam Tip: Notice how factoring reveals the structure of the answer. Look for patterns in your intermediate results - they often guide you toward the most elegant final form.

Question 12. Differentiate \( (x^3 \cos x - 2^x \tan x) \)
Answer: To find the derivative of \( x^3 \cos x - 2^x \tan x \), we differentiate each term separately using the product rule.

For the first term \( x^3 \cos x \), let \( u = x^3 \) and \( v = \cos x \). Then \( u' = 3x^2 \) and \( v' = -\sin x \).

\( [x^3 \cos x]' = 3x^2 \cos x - x^3 \sin x = x^2(3\cos x - x \sin x) \)

For the second term \( 2^x \tan x \), let \( u = 2^x \) and \( v = \tan x \). Then \( u' = 2^x \log 2 \) and \( v' = \sec^2 x \).

\( [2^x \tan x]' = 2^x \log 2 \tan x + 2^x \sec^2 x = 2^x(\log 2 \tan x + \sec^2 x) \)

Combining both:

\( \frac{d}{dx}[x^3 \cos x - 2^x \tan x] = x^2(3\cos x - x \sin x) - 2^x(\log 2 \tan x + \sec^2 x) \)

In simple words: Apply the product rule to each part independently, factor where possible, then subtract the derivative of the second term from the derivative of the first.

Exam Tip: When a function is a difference of two products, differentiate each product separately, then combine. Remember the derivative of \( a^x \) is \( a^x \log a \), not \( a^x \ln a \) unless specified.

Exercise 28D

Question 1. Differentiate \( \frac{2^x}{x} \)
Answer: To find the derivative of \( \frac{2^x}{x} \), we apply the quotient rule.

Let \( u = 2^x \) and \( v = x \).

Then \( u' = 2^x \log 2 \) and \( v' = 1 \).

Using the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \):

\( \left(\frac{2^x}{x}\right)' = \frac{2^x \log 2 \cdot x - 2^x \cdot 1}{x^2} = \frac{2^x(x \log 2 - 1)}{x^2} \)

In simple words: In the quotient rule, multiply the derivative of the top by the bottom, subtract the top times the derivative of the bottom, then divide by the square of the bottom.

Exam Tip: Always factor out common terms from the numerator to simplify. The derivative of \( 2^x \) uses the natural logarithm of the base: \( 2^x \log 2 \).

Question 2. Differentiate \( \frac{\log x}{x} \)
Answer: To find the derivative of \( \frac{\log x}{x} \), we apply the quotient rule.

Let \( u = \log x \) and \( v = x \).

Then \( u' = \frac{1}{x} \) and \( v' = 1 \).

Using the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \):

\( \left(\frac{\log x}{x}\right)' = \frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \)

In simple words: The derivative of the logarithm is \( \frac{1}{x} \). After applying the quotient rule and simplifying, the \( x \) terms in the numerator cancel out.

Exam Tip: Notice how the \( x \) in the numerator cancels with the reciprocal \( \frac{1}{x} \) during simplification. Always look for such cancellations to make your final answer cleaner.

Question 3. Differentiate \( \frac{e^x}{1 + x} \)
Answer: To find the derivative of \( \frac{e^x}{1 + x} \), we apply the quotient rule.

Let \( u = e^x \) and \( v = 1 + x \).

Then \( u' = e^x \) and \( v' = 1 \).

Using the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \):

\( \left(\frac{e^x}{1 + x}\right)' = \frac{e^x(1 + x) - e^x \cdot 1}{(1 + x)^2} = \frac{e^x[(1 + x) - 1]}{(1 + x)^2} = \frac{xe^x}{(1 + x)^2} \)

In simple words: Factor out \( e^x \) from the numerator after applying the quotient rule, and simplify the bracket to get just \( x \).

Exam Tip: Always try to factor out exponential terms from the numerator to simplify. The final form \( \frac{xe^x}{(1 + x)^2} \) is much cleaner than leaving it in expanded form.

Question 4. Differentiate \( \frac{e^x}{(1+x^2)} \)
Answer: To find the derivative of \( \frac{e^x}{(1+x^2)} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = e^x \) and \( v = (1+x^2) \). Then: \( u' = \frac{du}{dx} = \frac{d(e^x)}{dx} = e^x \) \( v' = \frac{dv}{dx} = \frac{d(1+x^2)}{dx} = 2x \) Applying the quotient rule formula: \( \left(\frac{e^x}{(1+x^2)}\right)' = \frac{e^x \times (1+x^2) - e^x \times 2x}{(1+x^2)^2} \) \( = \frac{e^x(1+x^2-2x)}{(1+x^2)^2} \) \( = \frac{e^x(x-1)^2}{(1+x^2)^2} \)
In simple words: To differentiate a fraction, keep the bottom part, multiply it by the derivative of the top part, then subtract the top part multiplied by the derivative of the bottom part. Finally, divide everything by the square of the bottom part.

Exam Tip: Always verify your numerator simplification - factoring expressions like \( x^2 - 2x + 1 \) into \( (x-1)^2 \) earns marks and shows clear understanding.

Question 5. Differentiate \( \left(\frac{2x^2-4}{3x^2+7}\right) \)
Answer: To find the derivative of \( \frac{2x^2-4}{3x^2+7} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (2x^2-4) \) and \( v = (3x^2+7) \). Then: \( u' = \frac{du}{dx} = \frac{d(2x^2-4)}{dx} = 4x \) \( v' = \frac{dv}{dx} = \frac{d(3x^2+7)}{dx} = 6x \) Applying the quotient rule formula: \( \left[\frac{(2x^2-4)}{(3x^2+7)}\right]' = \frac{4x \times (3x^2+7) - (2x^2-4) \times 6x}{(3x^2+7)^2} \) \( = \frac{12x^3 + 28x - 12x^3 + 24x}{(3x^2+7)^2} \) \( = \frac{52x}{(3x^2+7)^2} \)
In simple words: When you have two polynomials in a fraction, find the derivative of each part separately, then use the quotient rule - this means multiply each derivative by the opposite function and subtract them, putting the result over the square of the denominator.

Exam Tip: Watch for like terms that cancel when you expand the numerator - combining \( 28x \) and \( 24x \) correctly to get \( 52x \) demonstrates careful algebraic manipulation.

Question 6. Differentiate \( \left(\frac{x^2+3x-1}{x+2}\right) \)
Answer: To find the derivative of \( \frac{x^2+3x-1}{x+2} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (x^2+3x-1) \) and \( v = (x+2) \). Then: \( u' = \frac{du}{dx} = \frac{d(x^2+3x-1)}{dx} = 2x+3 \) \( v' = \frac{dv}{dx} = \frac{d(x+2)}{dx} = 1 \) Applying the quotient rule formula: \( \left(\frac{x^2+3x-1}{x+2}\right)' = \frac{(2x+3) \times (x+2) - (x^2+3x-1) \times 1}{(x+2)^2} \) \( = \frac{2x^2 + 4x + 3x + 6 - x^2 - 3x + 1}{(x+2)^2} \) \( = \frac{x^2 + 4x + 7}{(x+2)^2} \)
In simple words: Take the derivative of the top expression and multiply by the bottom, then subtract the top multiplied by the derivative of the bottom. Divide this entire result by the denominator squared.

Exam Tip: Expanding the numerator step-by-step and carefully combining like terms prevents errors - this is where most marks are lost.

Question 7. Differentiate \( \frac{(x^2-1)}{(x^2+7x+1)} \)
Answer: To find the derivative of \( \frac{x^2-1}{x^2+7x+1} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (x^2-1) \) and \( v = (x^2+7x+1) \). Then: \( u' = \frac{du}{dx} = \frac{d(x^2-1)}{dx} = 2x \) \( v' = \frac{dv}{dx} = \frac{d(x^2+7x+1)}{dx} = 2x+7 \) Applying the quotient rule formula: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) \( \left[\frac{(x^2-1)}{(x^2+7x+1)}\right]' = \frac{2x(x^2+7x+1) - (x^2-1)(2x+7)}{(x^2+7x+1)^2} \) \( = \frac{2x^3 + 14x^2 + 2x - 2x^3 - 7x^2 + 2x + 7}{(x^2+7x+1)^2} \) \( = \frac{7x^2 + 4x + 7}{(x^2+7x+1)^2} \)
In simple words: Differentiate the top part and the bottom part separately. Multiply the top's derivative by the whole bottom, then subtract the whole top times the bottom's derivative. Place everything over the denominator squared.

Exam Tip: Organize your expansion of the numerator in columns by degree - grouping \( x^3 \), \( x^2 \), \( x \), and constant terms makes cancellation obvious.

Question 8. Differentiate \( \left(\frac{5x^2+6x+7}{2x^2+3x+4}\right) \)
Answer: To find the derivative of \( \frac{5x^2+6x+7}{2x^2+3x+4} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (5x^2+6x+7) \) and \( v = (2x^2+3x+4) \). Then: \( u' = \frac{du}{dx} = \frac{d(5x^2+6x+7)}{dx} = 10x+6 \) \( v' = \frac{dv}{dx} = \frac{d(2x^2+3x+4)}{dx} = 4x+3 \) Applying the quotient rule formula: \( \left(\frac{5x^2+6x+7}{2x^2+3x+4}\right)' = \frac{(10x+6)(2x^2+3x+4) - (5x^2+6x+7)(4x+3)}{(2x^2+3x+4)^2} \) \( = \frac{20x^3 + 30x^2 + 40x + 12x^2 + 18x + 24 - 20x^3 - 15x^2 - 24x^2 - 18x - 28x - 21}{(2x^2+3x+4)^2} \) \( = \frac{3x^2 + 12x + 3}{(2x^2+3x+4)^2} \) \( = \frac{3(x^2+4x+1)}{(2x^2+3x+4)^2} \)
In simple words: Take the slope of the top times the whole bottom, subtract the whole top times the slope of the bottom, and divide by the denominator squared. Simplifying by factoring out common factors improves the final answer.

Exam Tip: Always check whether the numerator can be factored - pulling out a common factor like 3 shows thoroughness and often reveals the simplest form.

Question 9. Differentiate \( \frac{x}{(a^2+x^2)} \)
Answer: To find the derivative of \( \frac{x}{a^2+x^2} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = x \) and \( v = (a^2+x^2) \). Then: \( u' = \frac{du}{dx} = \frac{d(x)}{dx} = 1 \) \( v' = \frac{dv}{dx} = \frac{d(a^2+x^2)}{dx} = 2x \) Applying the quotient rule formula: \( \left[\frac{x}{(a^2+x^2)}\right]' = \frac{1 \times (a^2+x^2) - (x) \times (2x)}{(a^2+x^2)^2} \) \( = \frac{a^2 + x^2 - 2x^2}{(a^2+x^2)^2} \) \( = \frac{a^2 - x^2}{(a^2+x^2)^2} \)
In simple words: Differentiate the top to get 1, multiply it by the whole bottom to get \( a^2 + x^2 \). Then subtract the original top multiplied by the derivative of the bottom, which gives \( 2x^2 \). Place the result over the denominator squared.

Exam Tip: Notice how \( a \) is a constant - only the \( x^2 \) term changes. This simplification in the numerator is key to obtaining the clean final form.

Question 10. Differentiate \( \frac{x^4}{\sin x} \)
Answer: To find the derivative of \( \frac{x^4}{\sin x} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = x^4 \) and \( v = \sin x \). Then: \( u' = \frac{du}{dx} = \frac{d(x^4)}{dx} = 4x^3 \) \( v' = \frac{dv}{dx} = \frac{d(\sin x)}{dx} = \cos x \) Applying the quotient rule formula: \( \left[\frac{x^4}{\sin x}\right]' = \frac{4x^3 \times (\sin x) - (x^4) \times (\cos x)}{(\sin x)^2} \) \( = \frac{x^3[4(\sin x) - x(\cos x)]}{(\sin x)^2} \)
In simple words: Take the power rule to get \( 4x^3 \) for the numerator, and remember that the derivative of sine is cosine. Apply the quotient rule, then factor out the common \( x^3 \) from the numerator to simplify.

Exam Tip: Factoring out \( x^3 \) demonstrates algebraic insight - examiners value answers that show simplification beyond just applying the rule mechanically.

Question 11. Differentiate \( \frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}} \)
Answer: To find the derivative of \( \frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (\sqrt{a}+\sqrt{x}) \) and \( v = (\sqrt{a}-\sqrt{x}) \). Then: \( u' = \frac{du}{dx} = \frac{d(\sqrt{a}+\sqrt{x})}{dx} = \frac{1}{2\sqrt{x}} \) \( v' = \frac{dv}{dx} = \frac{d(\sqrt{a}-\sqrt{x})}{dx} = -\frac{1}{2\sqrt{x}} \) Applying the quotient rule formula: \( \left[\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\right]' = \frac{\frac{1}{2\sqrt{x}} \times (\sqrt{a}-\sqrt{x}) - (\sqrt{a}+\sqrt{x}) \times (-\frac{1}{2\sqrt{x}})}{(\sqrt{a}-\sqrt{x})^2} \) \( = \frac{\frac{\sqrt{a}}{2\sqrt{x}} - \frac{1}{2} + \frac{\sqrt{a}}{2\sqrt{x}} + \frac{1}{2}}{(\sqrt{a}-\sqrt{x})^2} \) \( = \frac{\frac{\sqrt{a}}{\sqrt{x}}}{(\sqrt{a}-\sqrt{x})^2} \) \( = \frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^2} \)
In simple words: Since \( a \) is constant, its derivative is zero. Only \( \sqrt{x} \) changes with derivative \( \frac{1}{2\sqrt{x}} \). After applying the quotient rule, the constant terms cancel nicely, leaving a compact expression.

Exam Tip: Pay attention to the signs when \( v' \) is negative - this causes the second term to flip sign when subtracted, leading to addition instead.

Question 12. Differentiate \( \frac{\cos x}{\log x} \)
Answer: To find the derivative of \( \frac{\cos x}{\log x} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = \cos x \) and \( v = \log x \). Recall the formulas: \( \frac{d(\cos x)}{dx} = -\sin x \) \( \frac{d(\log x)}{dx} = \frac{1}{x} \) Applying the quotient rule formula: \( \left[\frac{\cos x}{\log x}\right]' = \frac{(-\sin x)(\log x) - (\cos x)\left(\frac{1}{x}\right)}{(\log x)^2} \) \( = \frac{-x\sin x(\log x) - (\cos x)}{x(\log x)^2} \)
In simple words: The derivative of cosine is negative sine, and the derivative of natural log is one over x. Substitute these into the quotient rule - multiply each derivative by the opposite function, subtract, then divide by the logarithm squared.

Exam Tip: Watch for the negative sign in front of sine - this is a common source of error. Always double-check your derivative formulas for trigonometric functions.

Question 13. Differentiate \( \frac{2\cot x}{\sqrt{x}} \)
Answer: To find the derivative of \( \frac{2\cot x}{\sqrt{x}} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = 2\cot x \) and \( v = \sqrt{x} \). Then: \( u' = \frac{du}{dx} = \frac{d(2\cot x)}{dx} = -2\csc^2 x \) \( v' = \frac{dv}{dx} = \frac{d(\sqrt{x})}{dx} = \frac{1}{2\sqrt{x}} \) Applying the quotient rule formula: \( \left[\frac{2\cot x}{\sqrt{x}}\right]' = \frac{(-2\csc^2 x)(\sqrt{x}) - (2\cot x)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2} \) \( = \frac{-2x\csc^2 x - (\cot x)}{x} \) \( = \frac{-2x\csc^2 x - \cot x}{x^{3/2}} \)
In simple words: The derivative of \( 2\cot x \) is \( -2\csc^2 x \), and for the square root denominator we get \( \frac{1}{2\sqrt{x}} \). Apply the quotient rule - the numerator becomes products of these derivatives with the opposite functions, divided by the denominator squared.

Exam Tip: Remember that the derivative of cotangent involves a negative sign - this is frequently tested and often missed by students.

Question 14. Differentiate \( \frac{\sin x}{(1+\cos x)} \)
Answer: To find the derivative of \( \frac{\sin x}{1+\cos x} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = \sin x \) and \( v = (1+\cos x) \). Then: \( u' = \frac{du}{dx} = \frac{d(\sin x)}{dx} = \cos x \) \( v' = \frac{dv}{dx} = \frac{d(1+\cos x)}{dx} = -\sin x \) Applying the quotient rule formula: \( \left[\frac{\sin x}{(1+\cos x)}\right]' = \frac{\cos x(1+\cos x) - (\sin x)(-\sin x)}{(1+\cos x)^2} \) \( = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2} \) \( = \frac{\cos x + 1}{(1+\cos x)^2} \) \( = \frac{1}{(1+\cos x)} \)
In simple words: The derivative of sine is cosine, and the derivative of cosine is negative sine. After substituting into the quotient rule, use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the numerator. The final answer factors beautifully.

Exam Tip: Recognizing when to apply the Pythagorean identity is crucial - the numerator simplifies dramatically to \( \cos x + 1 \), which cancels with part of the denominator.

Question 15. Differentiate \( \left(\frac{1+\sin x}{1-\sin x}\right) \)
Answer: To find the derivative of \( \frac{1+\sin x}{1-\sin x} \), we use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (1+\sin x) \) and \( v = (1-\sin x) \). Then: \( u' = \frac{du}{dx} = \frac{d(1+\sin x)}{dx} = \cos x \) \( v' = \frac{dv}{dx} = \frac{d(1-\sin x)}{dx} = -\cos x \) Applying the quotient rule formula: \( \left[\frac{1+\sin x}{1-\sin x}\right]' = \frac{\cos x(1-\sin x) - (1+\sin x)(-\cos x)}{(1-\sin x)^2} \) \( = \frac{\cos x - \cos x \sin x + \cos x + \cos x \sin x}{(1-\sin x)^2} \) \( = \frac{2\cos x}{(1-\sin x)^2} \)
In simple words: Both the constant 1 and the sine function have derivatives; 1 gives 0 and sine gives cosine. Apply the quotient rule and notice that the cross terms \( -\cos x \sin x \) and \( +\cos x \sin x \) cancel, leaving only \( 2\cos x \) in the numerator.

Exam Tip: Watch for the cancellation of middle terms - this is the key insight that makes the final answer clean. Always expand fully before concluding.

Question 16. Differentiate \( \left(\frac{1-\cos x}{1+\cos x}\right) \)
Answer: To find the derivative of \( \frac{1-\cos x}{1+\cos x} \), we apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) where \( v \neq 0 \). Let \( u = (1-\cos x) \) and \( v = (1+\cos x) \). Then: \( u' = \frac{du}{dx} = \frac{d(1-\cos x)}{dx} = \sin x \) \( v' = \frac{dv}{dx} = \frac{d(1+\cos x)}{dx} = -\sin x \) Applying the quotient rule formula: \( \left[\frac{1-\cos x}{1+\cos x}\right]' = \frac{\sin x(1+\cos x) - (1-\cos x)(-\sin x)}{(1+\cos x)^2} \) \( = \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{(1+\cos x)^2} \) \( = \frac{2\sin x}{(1+\cos x)^2} \)
In simple words: Take the derivative of the numerator to get sine, and the derivative of the denominator to get negative sine. Substitute into the quotient rule - when you expand, the product terms \( \sin x \cos x \) cancel out, leaving \( 2\sin x \) in the numerator.

Exam Tip: This problem mirrors Question 15 but with cosine instead - spotting the parallel structure helps you work more confidently. The cancellation pattern is identical.

Question 17. Differentiate \( \left(\frac{\sin x - \cos x}{\sin x + \cos x}\right) \)
Answer: To find the derivative of this function, we take \( u = (\sin x - \cos x) \) and \( v = (\sin x + \cos x) \). Using the quotient rule, where \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), we determine:

\( u' = \frac{d(\sin x - \cos x)}{dx} = \cos x + \sin x \)

\( v' = \frac{d(\sin x + \cos x)}{dx} = \cos x - \sin x \)

Applying the quotient rule formula:

\( \left[\frac{\sin x - \cos x}{\sin x + \cos x}\right]' = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \)

\( = \frac{\sin^2 x + \cos^2 x + 2\sin x \cos x - (\sin x \cos x - \sin^2 x - \cos^2 x + \sin x \cos x)}{(\sin x + \cos x)^2} \)

\( = \frac{\sin^2 x + \cos^2 x + 2\sin x \cos x + \sin^2 x + \cos^2 x - 2\sin x \cos x}{(\sin x + \cos x)^2} \)

\( = \frac{2(\sin^2 x + \cos^2 x)}{\sin^2 x + \cos^2 x + 2\sin x \cos x} \)

\( = \frac{2}{1 + \sin 2x} \)
In simple words: Use the quotient rule to differentiate a fraction of trigonometric functions by taking the derivative of the numerator and denominator separately, then substitute into the formula.

Exam Tip: Always verify that \( \sin^2 x + \cos^2 x = 1 \) before simplifying, and remember the double angle identity \( 2\sin x \cos x = \sin 2x \).

Question 18. Differentiate \( \left(\frac{\sec x - \tan x}{\sec x + \tan x}\right) \)
Answer: To find this derivative, set \( u = (\sec x - \tan x) \) and \( v = (\sec x + \tan x) \). The quotient rule gives us:

\( u' = \frac{d(\sec x - \tan x)}{dx} = \sec x \tan x - \sec^2 x \)

\( v' = \frac{d(\sec x + \tan x)}{dx} = \sec x \tan x + \sec^2 x \)

Using the quotient rule:

\( \left[\frac{\sec x - \tan x}{\sec x + \tan x}\right]' = \frac{(\sec x \tan x - \sec^2 x)(\sec x + \tan x) - (\sec x - \tan x)(\sec x \tan x + \sec^2 x)}{(\sec x + \tan x)^2} \)

\( = \frac{(\sec x + \tan x)[(\sec x \tan x - \sec^2 x) - (\sec x - \tan x)(\sec x)]}{(\sec x + \tan x)^2} \)

\( = \frac{(\sec x + \tan x)[(\sec x \tan x - \sec^2 x) - (\sec^2 x - \sec x \tan x)]}{(\sec x + \tan x)^2} \)

\( = \frac{(\sec x + \tan x)[2\sec x \tan x - 2\sec^2 x]}{(\sec x + \tan x)^2} \)

\( = \frac{2\sec x(\tan x - \sec x)}{(\sec x + \tan x)} \)
In simple words: Apply the quotient rule to differentiate a fraction containing secant and tangent functions by finding the derivatives of both the top and bottom parts.

Exam Tip: Factor out common terms like \( \sec x \) from the numerator to simplify the final answer and avoid algebraic errors.

Question 19. Differentiate \( \left(\frac{e^x + \sin x}{1 + \log x}\right) \)
Answer: To find the derivative, we set \( u = (e^x + \sin x) \) and \( v = (1 + \log x) \). Using the quotient rule:

\( u' = \frac{d(e^x + \sin x)}{dx} = e^x + \cos x \)

\( v' = \frac{d(1 + \log x)}{dx} = \frac{1}{x} \)

Applying the quotient rule:

\( \left[\frac{e^x + \sin x}{1 + \log x}\right]' = \frac{(e^x + \cos x)(1 + \log x) - (e^x + \sin x)\left(\frac{1}{x}\right)}{(1 + \log x)^2} \)

\( = \frac{x(e^x + \cos x)(1 + \log x) - (e^x + \sin x)}{x(1 + \log x)^2} \)
In simple words: Remember that the derivative of \( e^x \) is \( e^x \) itself, the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \log x \) is \( \frac{1}{x} \).

Exam Tip: Carefully track all terms in the numerator when expanding products, and multiply through by x to clear fractions before presenting the final answer.

Question 20. Differentiate \( \left(\frac{e^x \sin x}{\sec x}\right) \)
Answer: To find this derivative, set \( u = (e^x \sin x) \) and \( v = (\sec x) \). First, find \( u' \) using the product rule:

For \( u = e^x \sin x \), with \( g = e^x \) and \( h = \sin x \):

\( u' = e^x \sin x + e^x \cos x \)

Next, find \( v' \):

\( v' = \sec x \tan x \)

Using the quotient rule:

\( \left[\frac{e^x \sin x}{\sec x}\right]' = \frac{(e^x \sin x + e^x \cos x)(\sec x) - (e^x \sin x)(\sec x \tan x)}{(\sec x)^2} \)

\( = \frac{(e^x \sin x + e^x \cos x) - (e^x \sin x)(\tan x)}{(\sec x)} \)
In simple words: When the numerator contains a product like \( e^x \sin x \), use the product rule to get its derivative before using the quotient rule.

Exam Tip: Always simplify by cancelling common factors like \( \sec x \) in both numerator and denominator after applying the quotient rule to make your final expression cleaner.

Question 21. Differentiate \( \left(\frac{2^x \cot x}{\sqrt{x}}\right) \)
Answer: To find this derivative, take \( u = (2^x \cot x) \) and \( v = (\sqrt{x}) \). First, find \( u' \) using the product rule with \( g = 2^x \) and \( h = \cot x \):

\( u' = (2^x \log 2)(\cot x) + 2^x(-\csc^2 x) = 2^x(\log 2 \cot x - \csc^2 x) \)

Next, find \( v' \):

\( v' = \frac{1}{2\sqrt{x}} \)

Using the quotient rule:

\( \left[\frac{2^x \cot x}{\sqrt{x}}\right]' = \frac{\{2^x[\log 2 \cot x - \csc^2 x] \times \sqrt{x}\} - \{(2^x \cot x) \times \frac{1}{2\sqrt{x}}\}}{x} \)

\( = \frac{\{2^x[\log 2 \cot x - \csc^2 x] \times \sqrt{x}\} - \{(2^{x-1} \cot x) \times \frac{1}{\sqrt{x}}\}}{x} \)

\( = \frac{\{2^x[x\log 2 \cot x - x\csc^2 x]\} - \{(2^{x-1} \cot x)\}}{x^{3/2}} \)
In simple words: The derivative of \( 2^x \) is \( 2^x \log 2 \), the derivative of \( \cot x \) is \( -\csc^2 x \), and the derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).

Exam Tip: Always apply the product rule to the numerator first when it contains a product, then use the quotient rule to complete the differentiation.

Question 22. Differentiate \( \left(\frac{e^x(x-1)}{(x+1)}\right) \)
Answer: To find this derivative, let \( u = e^x(x-1) \) and \( v = (x+1) \). First, find \( u' \) using the product rule with \( g = e^x \) and \( h = (x-1) \):

\( [e^x(x-1)]' = e^x(x-1) + e^x(1) = e^x x \)

Next, find \( v' \):

\( v' = 1 \)

Using the quotient rule:

\( \left[\frac{e^x(x-1)}{(x+1)}\right]' = \frac{(e^x x)(x+1) - [e^x(x-1)](1)}{(x+1)^2} \)

\( = \frac{e^x x^2 + e^x x - e^x x + e^x}{(x+1)^2} \)

\( = \frac{e^x x^2 + e^x}{(x+1)^2} \)

\( = \frac{e^x(x^2 + 1)}{(x+1)^2} \)
In simple words: Use the product rule to differentiate \( e^x(x-1) \), then apply the quotient rule with the denominator \( (x+1) \).

Exam Tip: After expanding the numerator, group and factor out the common exponential term to simplify the final expression.

Question 23. Differentiate \( \left(\frac{x \tan x}{(\sec x + \tan x)}\right) \)
Answer: To find this derivative, set \( u = (x \tan x) \) and \( v = (\sec x + \tan x) \). First, find \( u' \) using the product rule with \( g = x \) and \( h = \tan x \):

\( [x \tan x]' = (1)(\tan x) + x(\sec^2 x) = \tan x + x \sec^2 x \)

Next, find \( v' \):

\( v' = \sec x \tan x + \sec^2 x = \sec x(\tan x + \sec x) \)

Using the quotient rule:

\( \left[\frac{x \tan x}{(\sec x + \tan x)}\right]' = \frac{(\tan x + x\sec^2 x)(\sec x + \tan x) - [x \tan x][\sec x(\tan x + \sec x)]}{(\sec x + \tan x)^2} \)

\( = \frac{[\tan x + x\sec^2 x - x \tan x \sec x]}{(\sec x + \tan x)} \)

\( = \frac{\tan x + x \sec x(\sec x - \tan x)}{(\sec x + \tan x)} \)
In simple words: Use the product rule to differentiate the numerator \( x \tan x \), then apply the quotient rule to obtain the final derivative.

Exam Tip: Factor out \( \sec x \) from the denominator's derivative to simplify cancellations in the quotient rule formula.

Question 24. Differentiate \( \left(\frac{ax^2 + bx + c}{px^2 + qx + r}\right) \)
Answer: To find this derivative, let \( u = (ax^2 + bx + c) \) and \( v = (px^2 + qx + r) \). Calculate the derivatives:

\( u' = 2ax + b \)

\( v' = 2px + q \)

Using the quotient rule:

\( \left[\frac{ax^2 + bx + c}{px^2 + qx + r}\right]' = \frac{(2ax + b)(px^2 + qx + r) - (ax^2 + bx + c)(2px + q)}{(px^2 + qx + r)^2} \)

\( = \frac{2apx^3 + 2aqx^2 + 2arx + bpx^2 + bqx + br - [2apx^3 + aqx^2 + 2bpx^2 + bqx + 2cpx + cq]}{(px^2 + qx + r)^2} \)

\( = \frac{(aq - bp)x^2 + 2(ar - pc)x + br - cq}{(px^2 + qx + r)^2} \)
In simple words: For rational functions with polynomial numerators and denominators, the quotient rule produces a result where the numerator factors into \( (aq - bp)x^2 + 2(ar - pc)x + br - cq \).

Exam Tip: Expand both products in the quotient rule numerator completely, then collect like terms carefully to avoid sign errors.

Question 25. Differentiate \( \left(\frac{\sin x - x \cos x}{x \sin x + \cos x}\right) \)
Answer: To find this derivative, set \( u = (\sin x - x \cos x) \) and \( v = (x \sin x + \cos x) \). For \( u' \), use the product rule on the \( x \cos x \) term:

\( [x \cos x]' = (1)(\cos x) + x(-\sin x) = \cos x - x \sin x \)

So: \( u' = \cos x - (\cos x - x \sin x) = x \sin x \)

For \( v' \), use the product rule on the \( x \sin x \) term:

\( [x \sin x]' = (1)(\sin x) + x(\cos x) = \sin x + x \cos x \)

So: \( v' = (\sin x + x \cos x) - \sin x = x \cos x \)

Using the quotient rule:

\( \left[\frac{\sin x - x \cos x}{x \sin x + \cos x}\right]' = \frac{(x \sin x)(x \sin x + \cos x) - (\sin x - x \cos x)(x \cos x)}{(x \sin x + \cos x)^2} \)

\( = \frac{(x^2 \sin^2 x + x \sin x \cos x) - (x \sin x \cos x - x^2 \cos^2 x)}{(x \sin x + \cos x)^2} \)

\( = \frac{x^2(\sin^2 x + \cos^2 x)}{(x \sin x + \cos x)^2} \)

\( = \frac{x^2}{(x \sin x + \cos x)^2} \)
In simple words: Apply the product rule to each term containing a product of variables and trigonometric functions, then use the fundamental identity \( \sin^2 x + \cos^2 x = 1 \) to simplify.

Exam Tip: Always cancel middle terms when they appear with opposite signs after expanding, and use \( \sin^2 x + \cos^2 x = 1 \) to achieve a clean final answer.

Question 26. (i) Differentiate cotx
Answer: To find the derivative of \( \cot x \), use the quotient rule by writing \( \cot x = \frac{\cos x}{\sin x} \). Let \( u = \cos x \) and \( v = \sin x \):

\( u' = -\sin x \)

\( v' = \cos x \)

Using the quotient rule:

\( \frac{d(\cot x)}{dx} = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} \)

\( = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} \)

\( = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} \)

\( = \frac{-1}{\sin^2 x} \)

\( = -\csc^2 x \)
In simple words: Rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \), apply the quotient rule, and use the identity \( \sin^2 x + \cos^2 x = 1 \) to get \( -\csc^2 x \).

Exam Tip: Remember that the derivative of \( \cot x \) is negative, unlike the derivative of \( \tan x \), which is positive.

Question 26. (ii) Differentiate secx
Answer: To find the derivative of \( \sec x \), use the quotient rule by writing \( \sec x = \frac{1}{\cos x} \). Let \( u = 1 \) and \( v = \cos x \):

\( u' = 0 \)

\( v' = -\sin x \)

Using the quotient rule:

\( \frac{d(\sec x)}{dx} = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} \)

\( = \frac{\sin x}{\cos^2 x} \)

\( = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \)

\( = \tan x \sec x \)
In simple words: Express \( \sec x \) as \( \frac{1}{\cos x} \), use the quotient rule, and combine the resulting fractions using trigonometric identities to get \( \tan x \sec x \).

Exam Tip: Recognise that the final form \( \tan x \sec x \) is obtained by splitting the fraction \( \frac{\sin x}{\cos^2 x} \) into two complementary trigonometric ratios.

Question 1. Differentiate the following with respect to x: sin 4x
Answer: To find the derivative, apply the chain rule along with the differentiation formula for sine. Set y = sin 4x. Using the formula \( \frac{d}{dx}(\sin nu) = \cos(nu) \frac{d}{dx}(nu) \), we get \( \frac{d}{dx}(\sin 4x) = \cos(4x) \times \frac{d}{dx}(4x) = 4\cos 4x \). The derivative of y = sin 4x is 4cos 4x.
In simple words: When you differentiate sin 4x, bring down the number multiplying x (which is 4), change sine to cosine, and keep the 4x inside.

Exam Tip: Always remember the chain rule - multiply by the derivative of the inner function (here, the derivative of 4x is 4).

Question 2. Differentiate the following with respect to x: cos 5x
Answer: To find the derivative, use the chain rule with the differentiation formula for cosine. Let y = cos 5x. Using the formula \( \frac{d}{dx}(\cos nu) = - \sin(nu) \frac{d}{dx}(nu) \), we obtain \( \frac{d}{dx}(\cos 5x) = - \sin(5x) \times \frac{d}{dx}(5x) = - 5\sin 5x \). The derivative of y = cos 5x is - 5sin 5x.
In simple words: For cos 5x, bring down the coefficient 5, change cosine to negative sine, and keep 5x inside the function.

Exam Tip: The key difference from sine is that cosine derivatives carry a negative sign - don't forget this crucial detail.

Question 3. Differentiate the following with respect to x: tan 3x
Answer: To find the derivative, apply the chain rule with the differentiation formula for tangent. Set y = tan 3x. Using the formula \( \frac{d}{dx}(\tan nu) = \sec^2(nu) \frac{d}{dx}(nu) \), we get \( \frac{d}{dx}(\tan 3x) = \sec^2(3x) \times \frac{d}{dx}(3x) = 3\sec^2(3x) \). The derivative of y = tan 3x is 3sec²(3x).
In simple words: To differentiate tan 3x, multiply by 3, change tangent to sec squared, and keep 3x inside.

Exam Tip: Remember that the derivative of tangent uses sec² (not sec or any other function) - this is a common mistake to avoid.

Question 4. Differentiate the following with respect to x: cos x³
Answer: To find the derivative, use the chain rule twice since the exponent itself contains x. Let y = cos x³. Using the formulas \( \frac{d}{dx}(\cos nu) = - \sin nu \frac{d}{dx}(nu) \) and \( \frac{d}{dx}(x^n) = nx^{n-1} \), we have \( \frac{d}{dx}(\cos x^3) = - \sin(x^3) \times \frac{d}{dx}(x^3) = - 3x^2 \sin(x^3) \). The derivative of y = cos x³ is - 3x² sin(x³).
In simple words: The chain rule applies twice here - first bring down the exponent 3 to get 3x², then apply the negative sine, keeping x³ inside.

Exam Tip: Layered functions require applying the chain rule step by step from the outside in - don't skip any intermediate derivatives.

Question 5. Differentiate the following with respect to x: cot² x
Answer: To find the derivative, combine the chain rule with the product rule. Let y = cot² x. Using the formulas \( \frac{d}{dx}(\cot^a nu) = a\cot^{a-1}(nu) \times \frac{d}{dx}(\cot nu) \times \frac{d}{dx}(nu) \) and \( \frac{d}{dx}(x^n) = nx^{n-1} \), we get \( \frac{d}{dx}(\cot^2 x) = 2\cot(x) \times \frac{d\cot x}{dx} \times \frac{dx}{dx} = - 2\cot x (\csc^2 x) \). The derivative of y = cot² x is - 2cotx (cosec² x).
In simple words: Bring down the exponent 2 to get 2cot x, then multiply by the derivative of cotangent, which is negative cosec squared.

Exam Tip: When differentiating powers of trigonometric functions, always remember the negative sign that comes with the cosecant function.

Question 6. Differentiate the following with respect to x: tan³ x
Answer: To find the derivative, apply the chain rule to the power and then to the tangent function. Set y = tan³ x. Using the formula \( \frac{d}{dx}(\tan^n u) = n\tan^{n-1}(u) \times \frac{d(\tan u)}{dx} \times \frac{d(u)}{dx} \) with n = 3, we obtain \( \frac{d}{dx}(\tan^3 x) = 3\tan^2(x) \times \frac{d(\tan x)}{dx} = 3\tan^2 x \times (\sec^2 x) \). The derivative of y = tan³ x is 3tan² x × (sec² x).
In simple words: Bring down 3 as the power coefficient, reduce the exponent to 2, then multiply by sec² x (the derivative of tan x).

Exam Tip: For composite powers like tan³ x, apply the power rule first, then differentiate the base trigonometric function - two steps, not one.

Question 8. Differentiate the following with respect to x: e^(x²)
Answer: To find the derivative, use the chain rule with the exponential function. Let y = e^(x²). Using the formulas \( \frac{d}{dx}(e^a) = e^a \times \frac{d(a)}{dx} \) and \( \frac{d}{dx}(x^n) = nx^{n-1} \), we have \( \frac{d}{dx}(e^{x^2}) = e^{x^2} \times \frac{d}{dx}(x^2) = 2xe^{x^2} \). The derivative of y = e^(x²) is 2xe^(x²).
In simple words: When differentiating e to a power, keep e to that power and multiply by the derivative of the exponent.

Exam Tip: Exponential functions always remain in the final answer - never try to simplify e away. Just multiply by the derivative of what's in the exponent.

Question 9. Differentiate the following with respect to x: e^(cot x)
Answer: To find the derivative, apply the chain rule to the exponential with a trigonometric exponent. Let y = e^(cot x). Using the formula \( \frac{d}{dx}(e^a) = e^a \times \frac{d(a)}{dx} \), we obtain \( \frac{d}{dx}(e^{\cot x}) = e^{\cot x} \times \frac{d(\cot x)}{dx} = - e^{\cot x} \csc^2 x \). The derivative of y = e^(cot x) is - e^(cot x) cosec² x.
In simple words: Keep e to the cot x power, then multiply by the derivative of cot x, which is negative cosec squared.

Exam Tip: Never forget the negative sign that appears when differentiating cotangent - it's easy to miss and will cost you marks.

Question 10. Differentiate the following with respect to x: √(sin x)
Answer: To find the derivative, use nested chain rules for the square root and sine function. Let y = √(sin x). Using the formula \( \frac{d}{dx}(\sqrt{sinnu}) = \frac{1}{2\sqrt{sinnu}} \times \frac{d}{dx}(\sin u) \times \frac{d}{dx}(u) \), we get \( \frac{d}{dx}(\sqrt{\sin x}) = \frac{1}{2\sqrt{\sin x}} \times \frac{d}{dx}(\sin x) \times \frac{d}{dx}(x) = \frac{1}{2\sqrt{\sin x}} \times \cos x \). The derivative of y = √(sin x) is \( \frac{1}{2\sqrt{\sin x}} \) cos x.
In simple words: Pull down the fraction 1/2 with the square root in the denominator, then multiply by cos x (the derivative of sin x).

Exam Tip: Nested functions like square root of sine require careful layer-by-layer differentiation - write out each step to avoid algebraic errors.

Question 11. Differentiate the following with respect to x: (5 + 7x)⁶
Answer: To find the derivative, apply the chain rule to a polynomial raised to a power. Let y = (5 + 7x)⁶. Using the formula \( \frac{d}{dx}(y^n) = ny^{n-1} \times \frac{dy}{dx} \), we have \( \frac{d}{dx}(5 + 7x)^6 = 6(5 + 7x)^5 \times \frac{d}{dx}(5 + 7x) = 6(5 + 7x)^5 \times 7 = 42(5 + 7x)^5 \). The derivative of y = (5 + 7x)⁶ is 42(5 + 7x)⁵.
In simple words: Bring down 6 and reduce the power to 5, then multiply by 7 (the derivative of what's inside the parentheses).

Exam Tip: The chain rule requires multiplying by the derivative of the inner expression - forgetting this is the most frequent error on power function questions.

Question 12. Differentiate the following with respect to x: (3 - 4x)⁵
Answer: To find the derivative, use the chain rule on a power of a linear expression. Set y = (3 - 4x)⁵. Using the formula \( \frac{d}{dx}(y^n) = ny^{n-1} \times \frac{dy}{dx} \), we get \( \frac{d}{dx}(3 - 4x)^5 = 5(3 - 4x)^4 \times \frac{d}{dx}(3 - 4x) = 5(3 - 4x)^4 \times (-4) = - 20(3 - 4x)^4 \). The derivative of y = (3 - 4x)⁵ is - 16(3 - 4x)⁵.
In simple words: Lower the exponent from 5 to 4 and multiply by 5, then multiply by -4 (the derivative of the inner expression 3 - 4x).

Exam Tip: Watch for negative coefficients inside the parentheses - they must be included in your final answer to ensure correct signs.

Question 13. Differentiate the following with respect to x: (3x² - x + 1)⁴
Answer: To find the derivative, apply the chain rule to a quadratic polynomial raised to the fourth power. Let y = (3x² - x + 1)⁴. Using the formula \( \frac{d}{dx}(y^n) = ny^{n-1} \times \frac{dy}{dx} \), we obtain \( \frac{d}{dx}(3x^2 - x + 1)^4 = 4(3x^2 - x + 1)^3 \times \frac{d}{dx}(3x^2 - x + 1) = 4(3x^2 - x + 1)^3 \times (6x - 1) \). The derivative of y = (3x² - x + 1)⁴ is 4(3x² - x + 1)³(6x - 1).
In simple words: Bring down 4, reduce the power to 3, then multiply by the derivative of the polynomial inside, which is 6x - 1.

Exam Tip: Always differentiate the inner polynomial carefully - in this case, the derivative of 3x² - x + 1 is 6x - 1, not just any expression.

Question 14. Differentiate the following with respect to x: (ax² + bx + c)
Answer: To find the derivative of a general quadratic, differentiate each term separately using the power rule. Let y = ax² + bx + c. Using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), we have \( \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \). The derivative of y = (ax² + bx + c) is 2ax + b.
In simple words: Differentiate term by term - the x² term becomes 2ax, the x term becomes b, and the constant c becomes zero.

Exam Tip: General expressions with constants like a, b, c are treated just like numbered coefficients - apply the power rule to each term independently.

Question 15. Differentiate the following with respect to x: \( \frac{1}{(x^2 - x + 3)^3} \)
Answer: To find the derivative, rewrite the fraction as a negative power and apply the chain rule. Let y = (x² - x + 3)^(-3). Using the formula \( \frac{d}{dx}(y^n) = ny^{n-1} \times \frac{dy}{dx} \), we obtain \( \frac{d}{dx}(x^2 - x + 3)^{-3} = -3(x^2 - x + 3)^{-4} \times (2x - 1) = -3 \times \frac{1}{(x^2 - x + 3)^4} \times (2x - 1) \). The derivative of y = (x² - x + 3)^(-3) is \( \frac{-3(2x - 1)}{(x^2 - x + 3)^4} \).
In simple words: Rewrite the fraction as a negative power, bring down -3, reduce the power to -4, then multiply by the derivative of what's inside (which is 2x - 1).

Exam Tip: Converting reciprocals to negative exponents makes chain rule application much clearer - always use this strategy for fraction-type functions.

Question 16. Differentiate the following with respect to x: sin²(2x + 3)
Answer: To find the derivative, apply the chain rule twice - once for the square and once for the argument. Let y = sin²(2x + 3). Using the formula \( \frac{d}{dx}(\sin^2(ax + b)) = 2\sin(ax + b) \frac{d}{dx}(\sin(ax + b)) \frac{d}{dx}(ax + b) \), we have \( \frac{d}{dx}(\sin^2(2x + 3)) = 2\sin(2x + 3) \times \cos(2x + 3) \times 2 = 4\sin(2x + 3)\cos(2x + 3) \). The derivative of y = sin²(2x + 3) is 4sin(2x + 3)cos(2x + 3).
In simple words: Bring down 2 from the exponent, change one sine to cosine, keep the argument 2x + 3, then multiply by 2 (the derivative of 2x + 3).

Exam Tip: When differentiating squared trigonometric functions, you get both sine and cosine in your answer - this pattern always appears.

Question 17. Differentiate the following with respect to x: cos²(x³)
Answer: To find the derivative, apply nested chain rules for the square, cosine, and cubic functions. Let y = cos²(x³). Using the formula \( \frac{d}{dx}(\cos^a(nu)) = a\cos^{a-1}(nu) \frac{d}{dx}(\cos(nu)) \frac{d}{dx}(nu) \), we obtain \( \frac{d}{dx}(\cos^2(x^3)) = 2\cos(x^3) \times (-\sin(x^3)) \times 3x^2 = -6x^2\cos(x^3)\sin(x^3) \). The derivative of y = cos²(x³) is - 6x² cos(x³)sin x³.
In simple words: Lower the exponent from 2 to 1, change cosine to negative sine, keep x³ inside, then multiply by 3x² (the derivative of x³).

Exam Tip: Three levels of composition here - square, trig function, and polynomial - so three derivatives must be multiplied together in sequence.

Question 18. Differentiate the following with respect to x: √(sin x³)
Answer: To find the derivative, apply the chain rule to the square root, then to the sine, and finally to the cubic function. Let y = √(sin x³) = (sin x³)^(1/2). Using the formula \( \frac{d}{dx}(\sqrt{sinnu^a}) = \frac{1}{2\sqrt{sinnu^a}} \times \frac{d}{dx}(\sin(nu^a)) \times \frac{d}{dx}(nu^a) \), we get \( \frac{d}{dx}(\sqrt{\sin x^3}) = \frac{1}{2\sqrt{\sin x^3}} \times \cos(x^3) \times 3x^2 = \frac{3x^2\cos(x^3)}{2\sqrt{\sin x^3}} \). The derivative of y = √(sin x³) is \( \frac{3x^2(\cos x^3)}{2\sqrt{\sin x^3}} \).
In simple words: Pull 1/2 outside with the square root in the denominator, multiply by cos x³, then multiply by 3x² (the derivative of x³).

Exam Tip: Multiple layers of composition require extreme care - write each intermediate step to track where each derivative comes from.

Question 19. Differentiate the following with respect to x: √(x sin x)
Answer: To find the derivative, apply the chain rule to the square root, then use the product rule for the expression inside. Let y = √(x sin x) = (x sin x)^(1/2). Using the product rule inside along with the square root chain rule, we get \( \frac{d}{dx}(\sqrt{x\sin x}) = \frac{1}{2\sqrt{x\sin x}} \times \frac{d}{dx}(x\sin x) = \frac{1}{2\sqrt{x\sin x}} \times (\sin x + x\cos x) = \frac{\sin x + x\cos x}{2\sqrt{x\sin x}} \). The derivative of y = √(x sin x) is \( \frac{(\sin x + x\cos x)}{2\sqrt{x\sin x}} \).
In simple words: Use the product rule on x sin x to get sin x + x cos x, divide by 2√(x sin x), and simplify.

Exam Tip: When the square root contains a product, don't try to separate the square root - instead, differentiate the product as a whole first.

Question 20. Differentiate the following with respect to x: √(cot √x)
Answer: To find the derivative, apply three layers of chain rule for the square root, cotangent, and inner square root. Let y = √(cot √x) = (cot √x)^(1/2). Using the formula \( \frac{d}{dx}(\sqrt{cot\sqrt{x}}) = \frac{1}{2\sqrt{cot\sqrt{x}}} \times \frac{d}{dx}(\cot \sqrt{x}) \times \frac{d}{dx}(\sqrt{x}) \), we obtain \( \frac{d}{dx}(\sqrt{\cot\sqrt{x}}) = \frac{1}{2\sqrt{\cot\sqrt{x}}} \times (-\csc^2\sqrt{x}) \times \frac{1}{2\sqrt{x}} = \frac{-\sec^2\sqrt{x}}{4\sqrt{x}\sqrt{\cot\sqrt{x}}} \). The derivative of y = √(cot √x) is \( \frac{-\sec^2\sqrt{x}}{4\sqrt{x}\sqrt{\cot\sqrt{x}}} \).
In simple words: Work from outside in - pull 1/2 from the outer square root, multiply by negative cosec² from cotangent, then multiply by 1/(2√x) from the inner square root.

Exam Tip: Deeply nested functions demand systematic layer-by-layer work - number each stage if needed to prevent mixing up which derivative belongs where.

Question 21. Differentiate the following with respect to x: cos 3x sin 5x
Answer: Let \( y = \cos 3x \sin 5x \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = \sin 5x \frac{d(\cos 3x)}{dx} + \cos 3x \frac{d(\sin 5x)}{dx} \]
\[ = \sin 5x(-3\sin 3x) + \cos 3x(5\cos 5x) \]
\[ = 5\cos 3x \cos 5x - 3\sin 5x \sin 3x \]

Exam Tip: Always identify the two functions in a product and apply the product rule carefully - first times the derivative of second, plus second times the derivative of first.

Question 22. Differentiate the following with respect to x: sin x sin 2x
Answer: Let \( y = \sin x \sin 2x \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = \sin x \frac{d(\sin 2x)}{dx} + \sin 2x \frac{d(\sin x)}{dx} \]
\[ = \sin x(2\cos 2x) + \sin 2x(\cos x) \]
\[ = 2\sin x \cos 2x + \sin 2x \cos x \]

Exam Tip: Remember that the derivative of \( \sin nx \) is \( n \cos nx \), not just \( \cos nx \) - the chain rule multiplier is essential.

Question 23. Differentiate with respect to x: \( \cos(\sin\sqrt{ax + b}) \)
Answer: Let \( y = \cos(\sin\sqrt{ax + b}) \), \( z = \sin\sqrt{ax + b} \), and \( w = \sqrt{ax + b} \).
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dw} \times \frac{dw}{dx} \]
\[ = -\sin(\sin\sqrt{ax + b}) \times \cos\sqrt{ax + b} \times \frac{1}{2} \times (ax + b)^{-\frac{1}{2}} \times a \]
\[ = -\frac{a}{2}\sin(\sin\sqrt{ax + b}) \times \cos\sqrt{ax + b} \times (ax + b)^{-\frac{1}{2}} \]

Exam Tip: When dealing with nested functions, apply the chain rule step-by-step from the outermost function to the innermost - this prevents errors and ensures completeness.

Question 24. Differentiate with respect to x: \( e^{2x} \sin 3x \)
Answer: Let \( y = e^{2x} \sin 3x \), \( z = e^{2x} \), and \( w = \sin 3x \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = w \times \frac{dz}{dx} + z \times \frac{dw}{dx} \]
\[ = \sin 3x(2e^{2x}) + e^{2x}(3\cos 3x) \]
\[ = e^{2x}[2\sin 3x + 3\cos 3x] \]

Exam Tip: When exponential and trigonometric functions are multiplied, factor out the exponential term in your final answer for a cleaner form.

Question 25. Differentiate with respect to x: \( e^{3x} \cos 2x \)
Answer: Let \( y = e^{3x} \cos 2x \), \( z = e^{3x} \), and \( w = \cos 2x \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = w \times \frac{dz}{dx} + z \times \frac{dw}{dx} \]
\[ = \cos 2x(3e^{3x}) + e^{3x}(-2\sin 2x) \]
\[ = e^{3x}[3\cos 2x - 2\sin 2x] \]

Exam Tip: Be careful with negative signs when differentiating cosine - the derivative is negative sine, not positive.

Question 26. Differentiate with respect to x: \( e^{-5x} \cot 4x \)
Answer: Let \( y = e^{-5x} \cot 4x \), \( z = e^{-5x} \), and \( w = \cot 4x \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = w \times \frac{dz}{dx} + z \times \frac{dw}{dx} \]
\[ = \cot 4x(-5e^{-5x}) + e^{-5x}(-4\cosec^2 4x) \]
\[ = -e^{-5x}[5\cot 4x + 4\cosec^2 4x] \]

Exam Tip: The derivative of \( \cot u \) is \( -\cosec^2 u \times u' \) - remember both the negative sign and the chain rule multiplier.

Question 27. Differentiate with respect to x: \( \cos(x^3 \cdot e^x) \)
Answer: Let \( y = \cos(x^3 \cdot e^x) \), \( z = x^3 \cdot e^x \), \( m = e^x \), and \( w = x^3 \). Using the product rule,
\[ \frac{dz}{dx} = w \times \frac{dm}{dx} + m \times \frac{dw}{dx} \]
\[ = x^3(e^x) + e^x(3x^2) \]
\[ = e^x(x^3 + 3x^2) \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = -\sin(x^3 \cdot e^x) \times e^x(x^3 + 3x^2) \]

Exam Tip: For composite functions involving products inside trigonometric terms, first differentiate the inner product, then apply the chain rule with the trigonometric derivative.

Question 28. Differentiate with respect to x: \( e^{(x\sin x + \cos x)} \)
Answer: Let \( y = e^{(x\sin x + \cos x)} \), \( z = x\sin x + \cos x \), \( m = x \), and \( w = \sin x \). Using the product rule,
\[ \frac{dz}{dx} = w \times \frac{dm}{dx} + m \times \frac{dw}{dx} + \frac{d(\cos x)}{dx} \]
\[ = \sin x(1) + x(\cos x) - \sin x \]
\[ = x\cos x \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = e^{(x\sin x + \cos x)} \times x\cos x \]

Exam Tip: When differentiating the exponent that contains both products and individual terms, handle each piece carefully - the product rule applies only to the product portion.

Question 29. Differentiate with respect to x: \( \frac{e^x + e^{-x}}{e^x - e^{-x}} \)
Answer: Let \( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), \( u = e^x + e^{-x} \), and \( v = e^x - e^{-x} \). Using the quotient rule of differentiation,
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(e^x - e^{-x})(e^x - e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} \]
\[ = \frac{(e^x - e^{-x})^2 - (e^x + e^{-x})^2}{(e^x - e^{-x})^2} \]
Using the identity \( a^2 - b^2 = (a-b)(a+b) \),
\[ = \frac{(e^x - e^{-x} + e^x + e^{-x})(e^x - e^{-x} - e^x - e^{-x})}{(e^x - e^{-x})^2} \]
\[ = \frac{(2e^x)(-2e^{-x})}{(e^x - e^{-x})^2} \]
\[ = \frac{-4}{(e^x - e^{-x})^2} \]

Exam Tip: When simplifying quotients of exponential expressions, use algebraic identities like difference of squares to reduce the complexity significantly.

Question 30. Differentiate with respect to x: \( \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}} \)
Answer: Let \( y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}} \), \( u = e^{2x} + e^{-2x} \), and \( v = e^{2x} - e^{-2x} \). Using the quotient rule of differentiation,
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(e^{2x} - e^{-2x})(2e^{2x} - 2e^{-2x}) - (e^{2x} + e^{-2x})(2e^{2x} + 2e^{-2x})}{(e^{2x} - e^{-2x})^2} \]
\[ = \frac{2(e^{2x} - e^{-2x})^2 - 2(e^{2x} + e^{-2x})^2}{(e^{2x} - e^{-2x})^2} \]
Using the identity \( a^2 - b^2 = (a-b)(a+b) \),
\[ = \frac{2(e^{2x} - e^{-2x} + e^{2x} + e^{-2x})(e^{2x} - e^{-2x} - e^{2x} - e^{-2x})}{(e^{2x} - e^{-2x})^2} \]
\[ = \frac{2(2e^{2x})(-2e^{-2x})}{(e^{2x} - e^{-2x})^2} \]
\[ = \frac{-8}{(e^{2x} - e^{-2x})^2} \]

Exam Tip: The coefficients in the exponent (such as 2x instead of x) affect the derivatives proportionally - always account for these multipliers from the chain rule.

Question 31. Differentiate with respect to x: \( \sqrt{\frac{1-x^2}{1+x^2}} \)
Answer: Let \( y = \sqrt{\frac{1-x^2}{1+x^2}} \), \( u = 1 - x^2 \), \( v = 1 + x^2 \), and \( z = \frac{1-x^2}{1+x^2} \). Using the quotient rule of differentiation,
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]
\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} \]
\[ = \frac{-4x}{(1 + x^2)^2} \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2} \times \left(\frac{1-x^2}{1+x^2}\right)^{-\frac{1}{2}} \times \frac{-4x}{(1+x^2)^2} \]
\[ = \left[-2x(1-x^2)^{-\frac{1}{2}}\right] \times (1 + x^2)^{-\frac{3}{2}} \]

Exam Tip: For functions involving square roots of fractions, first simplify the fraction inside using the quotient rule, then apply the chain rule for the outer square root.

Question 32. Differentiate with respect to x: \( \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \)
Answer: Let \( y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \), \( u = a^2 - x^2 \), \( v = a^2 + x^2 \), and \( z = \frac{a^2-x^2}{a^2+x^2} \). Using the quotient rule of differentiation,
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(a^2 + x^2)(-2x) - (a^2 - x^2)(2x)}{(a^2 + x^2)^2} \]
\[ = \frac{-2xa^2 - 2x^3 - 2xa^2 + 2x^3}{(a^2 + x^2)^2} \]
\[ = \frac{-4xa^2}{(a^2 + x^2)^2} \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2} \times \left(\frac{a^2 - x^2}{a^2 + x^2}\right)^{-\frac{1}{2}} \times \frac{-4xa^2}{(a^2 + x^2)^2} \]
\[ = \left[-2xa^2(a^2 - x^2)^{-\frac{1}{2}}\right] \times (a^2 + x^2)^{-\frac{3}{2}} \]

Exam Tip: When constants like 'a' appear in the expression, treat them as you would any constant during differentiation - they affect the magnitude but not the structure of the derivative.

Question 33. Differentiate with respect to x: \( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \)
Answer: Let \( y = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \), \( u = 1 + \sin x \), \( v = 1 - \sin x \), and \( z = \frac{1+\sin x}{1-\sin x} \). Using the quotient rule of differentiation,
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1 - \sin x)(\cos x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2} \]
\[ = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{(1 - \sin x)^2} \]
\[ = \frac{2\cos x}{(1 - \sin x)^2} \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2} \times \left(\frac{1+\sin x}{1-\sin x}\right)^{-\frac{1}{2}} \times \frac{2\cos x}{(1-\sin x)^2} \]
\[ = \left[\cos x(1 + \sin x)^{-\frac{1}{2}}\right] \times (1 - \sin x)^{-\frac{3}{2}} \]

Exam Tip: Trigonometric expressions in denominators can simplify nicely - watch for terms that cancel when you combine the numerator terms carefully.

Question 34. Differentiate with respect to x: \( \sqrt{\frac{1 + e^x}{1 - e^x}} \)
Answer: Let \( y = \sqrt{\frac{1 + e^x}{1 - e^x}} \), \( u = 1 + e^x \), \( v = 1 - e^x \), and \( z = \frac{1+e^x}{1-e^x} \). Using the quotient rule of differentiation,
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1 - e^x)(e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2} \]
\[ = \frac{e^x - e^{2x} + e^x + e^{2x}}{(1 - e^x)^2} \]
\[ = \frac{2e^x}{(1 - e^x)^2} \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2} \times \left(\frac{1+e^x}{1-e^x}\right)^{-\frac{1}{2}} \times \frac{2e^x}{(1-e^x)^2} \]
\[ = \left[e^x(1 + e^x)^{-\frac{1}{2}}\right] \times (1 - e^x)^{-\frac{3}{2}} \]

Exam Tip: Exponential terms often cancel symmetrically in the numerator - pay close attention to sign changes when applying the quotient rule.

Question 35. Differentiate with respect to x: \( \frac{e^{2x} + x^3}{\cosec 2x} \)
Answer: Using the quotient rule of differentiation, if \( y = \frac{u}{v} \) where \( u = e^{2x} + x^3 \) and \( v = \cosec 2x \),
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{\cosec 2x(2e^{2x} + 3x^2) - (e^{2x} + x^3)(-2\cosec 2x \cot 2x)}{(\cosec 2x)^2} \]
\[ = \frac{2e^{2x}\cosec 2x + 3x^2 \cosec 2x + 2e^{2x} \cosec 2x \cot 2x + 2x^3 \cosec 2x \cot 2x}{(\cosec 2x)^2} \]
\[ = \frac{2e^{2x}\cosec 2x(1 + \cot 2x) + 3x^2 \cosec 2x(1 + \cot 2x)}{(\cosec 2x)^2} \]
\[ = \frac{(1 + \cot 2x)(2e^x + 3x^2)(\cosec 2x)}{(\cosec 2x)^2} \]
\[ = \frac{(1 + \cot 2x)(2e^x + 3x^2)}{\cosec 2x} \]

Exam Tip: When the denominator contains a trigonometric function, factor out common trigonometric terms to simplify - this reveals the structure and prevents calculation errors.

Question 36. Find \( \frac{dy}{dx} \) when \( y = \sin(\sqrt{\sin x + \cos x}) \)
Answer: Let \( y = \sin(\sqrt{\sin x + \cos x}) \) and \( z = \sqrt{\sin x + \cos x} \). Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \cos(\sqrt{\sin x + \cos x}) \times \frac{1}{2}(\sin x + \cos x)^{-\frac{1}{2}} \times (\cos x - \sin x) \]
\[ = \cos(\sin\sqrt{\sin x + \cos x}) \times \frac{1}{2}(\sin x + \cos x)^{-\frac{1}{2}} \times (\cos x - \sin x) \]

Exam Tip: Multiple layers of composition require careful application of the chain rule at each stage - work from the outermost function inward systematically.

Question 37. Find \( \frac{dy}{dx} \) when \( y = e^x \log(\sin 2x) \)
Answer: Let \( y = e^x \log(\sin 2x) \), \( z = e^x \), and \( w = \log(\sin 2x) \). Using the product rule of differentiation,
\[ \frac{dy}{dx} = w \times \frac{dz}{dx} + z \times \frac{dw}{dx} \]
\[ = \log(\sin 2x) \times e^x + e^x \times \frac{1}{\sin 2x} \times 2\cos 2x \]
\[ = e^x \times [\log(\sin 2x) + \frac{2\cos 2x}{\sin 2x}] \]
\[ = e^x \times [\log(\sin 2x) + 2\cot 2x] \]

Exam Tip: When differentiating logarithms of trigonometric functions, combine the result with other terms to express the final answer in its simplest form using trigonometric identities.

Question 38. Find \( \frac{dy}{dx} \) when \( y = \cos\left(\frac{1 - x^2}{1 + x^2}\right) \)
Answer: Let \( y = \cos\left(\frac{1 - x^2}{1 + x^2}\right) \), \( u = 1 - x^2 \), \( v = 1 + x^2 \), and \( z = \frac{1-x^2}{1+x^2} \). Using the quotient rule of differentiation,
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]
\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} \]
\[ = \frac{-4x}{(1 + x^2)^2} \]
Using the chain rule of differentiation,
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = -\sin\left(\frac{1 - x^2}{1 + x^2}\right) \times \frac{-4x}{(1 + x^2)^2} \]
\[ = \frac{4x\sin\left(\frac{1 - x^2}{1 + x^2}\right)}{(1 + x^2)^2} \]

Exam Tip: Ensure that when cancellations occur during quotient rule simplification, all terms are properly combined - leaving out a term will lead to an incorrect derivative.

Question 39. Find \( \frac{dy}{dx} \) when \( y = \sin\left(\frac{1+x^2}{1-x^2}\right) \)
Answer: Let \( y = \sin\left(\frac{1+x^2}{1-x^2}\right) \), \( u = 1 + x^2 \), \( v = 1 - x^2 \), \( z = \frac{1+x^2}{1-x^2} \)

Using the formulas \( \frac{d(x^2)}{dx} = 2x \) and \( \frac{d(\sin x)}{dx} = \cos x \)

By the quotient rule of differentiation, if \( z = \frac{u}{v} \):
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1-x^2)(2x) - (1+x^2)(-2x)}{(1-x^2)^2} \]
\[ = \frac{2x - 2x^3 + 2x + 2x^3}{(1-x^2)^2} \]
\[ = \frac{4x}{(1-x^2)^2} \]

By the chain rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \left[\cos\frac{1-x^2}{1+x^2}\right] \times \left[\frac{4x}{(1+x^2)^2}\right] \]
\[ = \left[\sin\frac{1-x^2}{1+x^2}\right] \times \left[\frac{4x}{(1+x^2)^2}\right] \]
In simple words: To find the derivative of a sine function with a fraction inside, use the chain rule - multiply the derivative of the sine part by the derivative of the inner fraction part.

Exam Tip: Always identify composite functions and apply the chain rule systematically; label intermediate variables to track each differentiation step clearly.

Question 40. Find \( \frac{dy}{dx} \) when \( y = \frac{\sin x + x^2}{\cot 2x} \)
Answer: Let \( y = \frac{\sin x + x^2}{\cot 2x} \), \( u = \sin x + x^2 \), \( v = \cot 2x \)

Using the formulas \( \frac{d(\sin x)}{dx} = \cos x \), \( \frac{d(x^n)}{dx} = n \times x^{n-1} \), and \( \frac{d(\cot x)}{dx} = -\csc^2 x \)

By the quotient rule of differentiation, if \( y = \frac{u}{v} \):
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{\cot 2x (\cos x + 2x) - (\sin x + x^2)(-2\csc^2 2x)}{(\cot 2x)^2} \]
\[ = \frac{\cot 2x (\cos x + 2x) + 2(\sin x + x^2)\csc^2 2x}{(\csc 2x)^2} \]
\[ = \frac{\cot 2x (\cos x + 2x) + 2\csc^2 2x (\sin x + x^2)}{(\csc 2x)^2} \]
\[ = \frac{2\csc^2 2x (\sin x + x^2)}{(\csc 2x)^2} + \frac{\cot 2x (\cos x + 2x)}{(\csc 2x)^2} \]
\[ = 2(\sin x + x^2) + \frac{\cos 2x (\cos x + 2x)}{\sin 2x \cdot \frac{1}{\sin^2 2x}} \]
\[ = 2(\sin x + x^2) + \cos 2x \sin 2x (\cos x + 2x) \]
In simple words: When differentiating a fraction with trigonometric functions, apply the quotient rule by taking the bottom times the derivative of the top, minus the top times the derivative of the bottom, then divide by the bottom squared.

Exam Tip: Track all trigonometric derivatives carefully - cosecant squared has a negative sign, and always simplify the final expression fully.

Question 41. If \( y = \frac{\cos x - \sin x}{\cos x + \sin x} \), show that \( \frac{dy}{dx} + y^2 + 1 = 0 \)
Answer: Let \( y = \frac{\cos x - \sin x}{\cos x + \sin x} \)

Using the formulas \( \frac{d(\sin x)}{dx} = \cos x \) and \( \frac{d(\cos x)}{dx} = -\sin x \)

By the quotient rule of differentiation, if \( y = \frac{u}{v} \):
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(\cos x + \sin x)(-\sin x - \cos x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2} \]
\[ = \frac{-(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x + \sin x)^2} \]
\[ = -\frac{(\cos x + \sin x)^2}{(\cos x + \sin x)^2} - \frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2} \]
\[ = -1 - y^2 \]

Therefore:
\[ \frac{dy}{dx} = -1 - y^2 \]
\[ \frac{dy}{dx} + y^2 + 1 = 0 \]
In simple words: We differentiate the given fraction using quotient rule, then show that the result equals negative one minus y squared, which confirms the required identity.

Exam Tip: When proving identities, carefully expand squared binomials and simplify; rewrite the final result in the form requested by the question to demonstrate the proof completely.

Question 42. If \( y = \frac{\cos x + \sin x}{\cos x - \sin x} \), show that \( \frac{dy}{dx} = \sec^2\left(x + \frac{\pi}{4}\right) \)
Answer: Let \( y = \frac{\cos x + \sin x}{\cos x - \sin x} \), \( u = \cos x + \sin x \), \( v = \cos x - \sin x \)

Using the formulas \( \frac{d(\sin x)}{dx} = \cos x \) and \( \frac{d(\cos x)}{dx} = -\sin x \)

By the quotient rule of differentiation, if \( y = \frac{u}{v} \):
\[ \frac{dy}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(\cos x - \sin x)(-\sin x + \cos x) - (\cos x + \sin x)(-\sin x - \cos x)}{(\cos x - \sin x)^2} \]
\[ = \frac{(\cos x - \sin x)^2 + (\cos x + \sin x)^2}{(\cos x - \sin x)^2} \]
\[ = \frac{(\cos^2 x + \sin^2 x - 2\cos x \sin x) + (\cos^2 x + \sin^2 x + 2\cos x \sin x)}{(\cos x - \sin x)^2} \]
\[ = \frac{2(\cos^2 x + \sin^2 x)}{(\cos x - \sin x)^2} \]
\[ = \frac{2}{(\cos x - \sin x)^2} \]
\[ = \frac{1}{\left(\frac{\cos x - \sin x}{\sqrt{2}}\right)^2} \]
\[ = \frac{1}{\left(\frac{\cos x \cos 45° - \sin x \sin 45°}{1}\right)^2} \]
\[ = \frac{1}{\cos^2(x+\frac{\pi}{4})} \]
\[ = \sec^2\left(x + \frac{\pi}{4}\right) \]
In simple words: Differentiate the given fraction using the quotient rule, simplify to get 2 over a squared expression, then rewrite the denominator as a cosine squared of a combined angle using the cosine addition formula.

Exam Tip: Recognize when to apply the identity \( \cos^2 x + \sin^2 x = 1 \) and when to use angle addition formulas - these conversions are key to reaching the required form.

Question 43. If \( y = \sqrt{\frac{1-x}{1+x}} \), prove that \( (1 - x^2)\frac{dy}{dx} + y = 0 \)
Answer: Let \( y = \sqrt{\frac{1-x}{1+x}} \), \( u = 1 - x \), \( v = 1 + x \), \( z = \frac{1-x}{1+x} \)

Using the formula \( \frac{d(x^1)}{dx} = 1 \)

By the quotient rule of differentiation, if \( z = \frac{u}{v} \):
\[ \frac{dz}{dx} = \frac{v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^2} \]
\[ = \frac{(1 + x)(-1) - (1 - x)(1)}{(1 + x)^2} \]
\[ = \frac{-1 - x - 1 + x}{(1 + x)^2} \]
\[ = \frac{-2}{(1 + x)^2} \]

By the chain rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2}\left(\frac{1-x}{1+x}\right)^{-\frac{1}{2}} \times \frac{-2}{(1+x)^2} \]
\[ = \frac{1}{2} \times \frac{(1+x)^{\frac{1}{2}}}{(1-x)^{\frac{1}{2}}} \times \frac{-2}{(1+x)^2} \]
\[ = -1 \times \frac{(1-x)^{-\frac{1}{2}}}{(1+x)^{\frac{3}{2}}} \times \frac{1}{(1+x)^1} \times \frac{1-x}{1-x} \]
\[ = -1 \times \frac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{3}{2}}} \times \frac{1}{(1-x)(1+x)} \]
\[ = -\frac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}} \times \frac{1}{(1-x)(1+x)} = -\frac{y}{1-x^2} \]

Therefore:
\[ (1 - x^2)\frac{dy}{dx} = -y \]
\[ (1 - x^2)\frac{dy}{dx} + y = 0 \]
In simple words: Find the derivative of the inner fraction using the quotient rule, then use the chain rule to multiply by the derivative of the square root, and the result simplifies to show the required relationship.

Exam Tip: When dealing with nested functions (square root of a fraction), always use substitution to label intermediate variables - this prevents algebraic errors and makes simplification clearer.

Question 44. If \( y = \sqrt{\frac{\sec x - \tan x}{\sec x + \tan x}} \), show that \( \frac{dy}{dx} = \sec x(\tan x + \sec x) \)
Answer: We have \( y = \sqrt{\frac{\sec x - \tan x}{\sec x + \tan x}} \)

Rewrite as:
\[ y = \sqrt{\frac{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}} = \sqrt{\frac{1 - \sin x}{1 + \sin x}} \]

Let \( u = 1 - \sin x \), \( v = 1 + \sin x \), \( z = \frac{1-\sin x}{1+\sin x} \)

Using the formula \( \frac{d(\sin x)}{dx} = \cos x \)

By the quotient rule of differentiation, if \( z = \frac{u}{v} \):
\[ \frac{dz}{dx} = \frac{(1 + \sin x)(-\cos x) - (1 - \sin x)(\cos x)}{(1 + \sin x)^2} \]
\[ = \frac{-\cos x - \sin x \cos x - \cos x + \sin x \cos x}{(1 + \sin x)^2} \]
\[ = \frac{-2\cos x}{(1 + \sin x)^2} \]

By the chain rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} \]
\[ = \frac{1}{2}\left(\frac{1-\sin x}{1+\sin x}\right)^{-\frac{1}{2}} \times \frac{-2\cos x}{(1+\sin x)^2} \]
\[ = -\frac{\cos x}{1} \times \left(\frac{1-\sin x}{1+\sin x}\right)^{-\frac{1}{2}} \times \frac{1}{(1+\sin x)^2} \]
\[ = \cos x \times (1 + \sin x)^{-\frac{1}{2}} \times (1 - \sin x)^{\frac{1}{2}} \times (1 - \sin x)^{-\frac{1}{2}} \times \left(\frac{1+\sin x}{1+\sin x}\right)^{\frac{3}{2}} \]

(Multiplying and dividing by \( (1 + \sin x)^{\frac{3}{2}} \))
\[ = \cos x \times (1 + \sin x)^{-\frac{1}{2}} \times \frac{1}{(1-x)(1+x)} \]
\[ = \cos x \times (1 + \sin x)^{-1} \times \frac{1-\sin x}{1-\sin x} \]
\[ = -\frac{y}{1-\sin^2 x} = -\frac{y}{\cos^2 x} \]
\[ = \frac{1+\sin x}{\cos^2 x} \]
\[ = \frac{1}{\cos x} \times \left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) \]
\[ = \sec x(\sec x + \tan x) \]
In simple words: Convert secant and tangent functions to their sine and cosine forms, then simplify the fraction inside the square root before differentiating using the quotient rule and chain rule together.

Exam Tip: Always simplify complex trigonometric expressions before differentiating - converting to basic trigonometric ratios often reveals hidden patterns that make the algebra much simpler.