NCERT Solutions Class 11 Mathematics Chapter 13 Limits and Derivatives

Exercise 13.1

 

Question 1. Evaluate the Given limit: \( \lim_{x \to 3} x + 3 \)
Answer: \( \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \)
In simple words: To find the limit, simply substitute the value 3 in place of x in the expression. Adding 3 and 3 gives 6.

Exam Tip: For simple polynomial limits where direct substitution does not lead to an indeterminate form like 0/0, directly plug in the limit value to get the answer quickly.

 

Question 2. Evaluate the Given limit: \( \lim_{x \to \pi} \left( x - \frac{22}{7} \right) \)
Answer: \( \lim_{x \to \pi} \left( x - \frac{22}{7} \right) = \left( \pi - \frac{22}{7} \right) \)
In simple words: Replace the variable x with the symbol \( \pi \) to calculate the limit. Since \( \pi \) and \( \frac{22}{7} \) are different, keep the answer in this subtractive format.

Exam Tip: Do not replace \( \pi \) with \( \frac{22}{7} \) unless specified, as \( \pi \) is an irrational number and \( \frac{22}{7} \) is only its rational approximation.

 

Question 3. Evaluate the Given limit: \( \lim_{r \to 1} \pi r^2 \)
Answer: \( \lim_{r \to 1} \pi r^2 = \pi (1)^2 = \pi \)
In simple words: Put the value 1 in place of r in the area formula. Since 1 squared is just 1, the final value is \( \pi \).

Exam Tip: When dealing with constant multipliers like \( \pi \), keep them outside the limit evaluation process to avoid any calculation errors.

 

Question 4. Evaluate the Given limit: \( \lim_{x \to 4} \frac{4x + 3}{x - 2} \)
Answer: \( \lim_{x \to 4} \frac{4x + 3}{x - 2} = \frac{4(4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2} \)
In simple words: Put the value 4 into the numerator and denominator. Work out the top and bottom parts separately to get the final fraction of 19 over 2.

Exam Tip: Always check if direct substitution makes the denominator zero. Since the denominator is non-zero here, substitution is the direct path to the solution.

 

Question 5. Evaluate the Given limit: \( \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} \)
Answer: \( \lim_{x \to -1} \frac{x^{10} + x^5 + 1}{x - 1} = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} = \frac{1 - 1 + 1}{-2} = -\frac{1}{2} \)
In simple words: Substitute -1 for x. Remember that a negative number raised to an even power becomes positive, while an odd power stays negative. Simplify to get minus 1/2.

Exam Tip: Be extra cautious with signs when raising negative numbers to odd or even powers, as even powers like 10 turn negative values positive, whereas odd powers like 5 retain the negative sign.

 

Question 6. Evaluate the Given limit: \( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} \)
Answer: We substitute \( x + 1 = y \), meaning \( y \to 1 \) when \( x \to 0 \).
Consequently,
\( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} = \lim_{y \to 1} \frac{y^5 - 1}{y - 1} \)
\( = \lim_{y \to 1} \frac{y^5 - 1^5}{y - 1} \)
\( = 5 \cdot 1^{5 - 1} \) [by applying the standard limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \)]
\( = 5 \)
Hence, \( \lim_{x \to 0} \frac{(x + 1)^5 - 1}{x} = 5 \)
In simple words: Since direct substitution gives 0/0, we change the variable by setting y = x + 1. This converts the expression into a standard limit formula, giving the answer 5.

Exam Tip: Substituting variables to convert complex expressions into standard algebraic limit forms is a very useful technique to solve indeterminate limits quickly.

 

Question 7. Evaluate the Given limit: \( \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} \)
Answer: When \( x = 2 \), the value of this rational function yields the indeterminate form \( \frac{0}{0} \).
Thus,
\( \lim_{x \to 2} \frac{3x^2 - x - 10}{x^2 - 4} = \lim_{x \to 2} \frac{(x - 2)(3x + 5)}{(x - 2)(x + 2)} \)
\( = \lim_{x \to 2} \frac{3x + 5}{x + 2} \)
\( = \frac{3(2) + 5}{2 + 2} \)
\( = \frac{11}{4} \)
In simple words: Substituting 2 directly gives 0/0. To solve this, factor the numerator and denominator, cancel out the common bracket (x - 2), and then plug in 2.

Exam Tip: If direct substitution yields \( \frac{0}{0} \), factorize the quadratic polynomial expressions to find and cancel the common factor causing the zero value.

 

Question 8. Evaluate the Given limit: \( \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} \)
Answer: When \( x = 3 \), the value of this rational function yields the indeterminate form \( \frac{0}{0} \).
Thus,
\( \lim_{x \to 3} \frac{x^4 - 81}{2x^2 - 5x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)} \)
\( = \lim_{x \to 3} \frac{(x + 3)(x^2 + 9)}{2x + 1} \)
\( = \frac{(3 + 3)(3^2 + 9)}{2(3) + 1} \)
\( = \frac{6 \times 18}{7} \)
\( = \frac{108}{7} \)
In simple words: Since putting 3 in gives 0/0, factorize both top and bottom. Cancel the common (x - 3) factor, then substitute 3 to calculate the final answer.

Exam Tip: Use the algebraic identity \( a^2 - b^2 = (a - b)(a + b) \) iteratively to split terms like \( x^4 - 81 \) into simpler factors.

 

Question 9. Evaluate the Given limit: \( \lim_{x \to 0} \frac{ax + b}{cx + 1} \)
Answer: By substituting \( x = 0 \) directly into the given limit, we get:
\( \lim_{x \to 0} \frac{ax + b}{cx + 1} = \frac{a(0) + b}{c(0) + 1} = \frac{b}{1} = b \)
In simple words: Simply put 0 in place of x. This gives b divided by 1, which simplifies directly to b.

Exam Tip: Always look for quick substitutions first. Since direct substitution does not yield \( \frac{0}{0} \) here, there is no need for any complex factoring or algebraic manipulation.

 

Question 10. Evaluate the Given limit: \( \lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} \)
Answer: When we evaluate at \( z = 1 \), the value of this rational expression results in the indeterminate form \( \frac{0}{0} \).
We substitute \( z^{\frac{1}{6}} = x \), which means that \( z \to 1 \) when \( x \to 1 \).
Therefore,
\( \lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} \)
\( = \lim_{x \to 1} \frac{x^2 - 1^2}{x - 1} \)
\( = 2 \cdot 1^{2 - 1} \) [using the standard limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \)]
\( = 2 \)
Hence, \( \lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = 2 \)
In simple words: Since putting in 1 gives 0/0, we change the variable by setting \( x = z^{\frac{1}{6}} \). This simplifies the exponents and lets us use a standard formula to get 2.

Exam Tip: Substituting fractional indices with a new variable makes rational limits much simpler to solve and prevents calculation errors with fractional powers.

 

Question 11. Evaluate the Given limit: \( \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} , a+b+c \neq 0 \)
Answer: Directly substitute \( x = 1 \) into the equation because the denominator is not equal to zero:
\( \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} = \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} \)
\( = \frac{a + b + c}{c + b + a} \)
\( = 1 \) [since \( a + b + c \neq 0 \)]
In simple words: Replace every x with 1. Since the top and the bottom parts of the fraction become exactly the same non-zero expression, dividing them gives 1.

Exam Tip: The condition \( a+b+c \neq 0 \) is crucial because it ensures we are not dividing zero by zero, which validates direct substitution.

 

Question 12. Evaluate the Given limit: \( \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} \)
Answer: When \( x = -2 \), the rational expression yields the indeterminate form \( \frac{0}{0} \).
By simplifying the numerator first, we get:
\( \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = \lim_{x \to -2} \frac{\left( \frac{2 + x}{2x} \right)}{x + 2} \)
\( = \lim_{x \to -2} \frac{2 + x}{2x(x + 2)} \)
\( = \lim_{x \to -2} \frac{1}{2x} \)
\( = \frac{1}{2(-2)} \)
\( = -\frac{1}{4} \)
In simple words: Putting in -2 directly gives 0/0. We simplify the fractions on top to combine them into one term, which allows us to cancel out the troublesome (x + 2) bracket. Then we substitute -2 to find the answer.

Exam Tip: For fractions within limits, combine the rational parts in the numerator into a single fraction first; this usually reveals common factors that can be cancelled.

 

Question 13. Evaluate the Given limit: \( \lim_{x \to 0} \frac{\sin ax}{bx} \)
Answer: Substituting \( x = 0 \) directly leads to the indeterminate form \( \frac{0}{0} \).
We can rewrite the expression as:
\( \lim_{x \to 0} \frac{\sin ax}{bx} = \lim_{x \to 0} \frac{\sin ax}{ax} \times \frac{ax}{bx} \)
\( = \lim_{x \to 0} \left( \frac{\sin ax}{ax} \right) \times \left( \frac{a}{b} \right) \)
\( = \frac{a}{b} \lim_{ax \to 0} \left( \frac{\sin ax}{ax} \right) \) [since \( x \to 0 \) implies \( ax \to 0 \)]
\( = \frac{a}{b} \times 1 \) [using the standard limit \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \)]
\( = \frac{a}{b} \)
In simple words: Since we have sin(ax) on top, we multiply the bottom by 'a' and divide by 'a' to match the angle. This matches the standard rule where sin(angle)/angle goes to 1, leaving us with a/b.

Exam Tip: For trigonometric limits involving sine, manipulate the denominator so that it perfectly matches the argument of the sine function to apply the standard formula.

 

Question 14. Evaluate the Given limit: \( \lim_{x \to 0} \frac{\sin ax}{\sin bx} , a, b \neq 0 \)
Answer: When \( x = 0 \), the given limit results in the indeterminate form \( \frac{0}{0} \).
We can reformulate the limit by dividing both the top and bottom expressions:
\( \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \lim_{x \to 0} \frac{\left( \frac{\sin ax}{ax} \right) \times ax}{\left( \frac{\sin bx}{bx} \right) \times bx} \)
\( = \left( \frac{a}{b} \right) \times \frac{\lim_{ax \to 0} \left( \frac{\sin ax}{ax} \right)}{\lim_{bx \to 0} \left( \frac{\sin bx}{bx} \right)} \) [since \( x \to 0 \) implies \( ax \to 0 \) and \( bx \to 0 \)]
\( = \left( \frac{a}{b} \right) \times \frac{1}{1} \) [using the identity \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \)]
\( = \frac{a}{b} \)
In simple words: Divide the numerator and denominator by x and adjust with constant factors so both parts match the standard sine limit format. Both sines become 1, leaving the final coefficient ratio a/b.

Exam Tip: When evaluating limits with sines on both top and bottom, divide both parts by x to isolate individual standard limits of the form \( \frac{\sin \theta}{\theta} \).

 

Question 15. Evaluate the Given limit: \( \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)} \)
Answer: We can observe that as \( x \) approaches \( \pi \), the term \( (\pi - x) \) approaches \( 0 \).
Therefore,
\( \lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)} = \frac{1}{\pi} \lim_{(\pi - x) \to 0} \frac{\sin(\pi - x)}{(\pi - x)} \)
\( = \frac{1}{\pi} \times 1 \) [using the standard limit \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \)]
\( = \frac{1}{\pi} \)
In simple words: Since x goes to \( \pi \), the value \( (\pi - x) \) goes to 0. This matches our standard sine limit template exactly, so that whole part becomes 1, leaving 1 over \( \pi \).

Exam Tip: Be sure to write the change of limit variable (e.g., \( \theta = \pi - x \to 0 \)) clearly on your paper to show the examiner you are applying the standard identity correctly.

 

Question 16. Evaluate the given limit: \( \lim_{x \to 0} \frac{\cos x}{\pi - x} \)
Answer: Plugging in \( x = 0 \) directly into the rational expression yields:
\( \lim_{x \to 0} \frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi} \)
In simple words: Since putting in 0 does not divide by zero, we can directly plug in 0. The cosine of 0 is 1, and the bottom is just \( \pi \), giving us 1 over \( \pi \).

Exam Tip: Do not use standard trigonometric limit laws like \( \lim \frac{\sin x}{x} = 1 \) for cosine expressions where the limit is well-defined and non-zero under direct substitution.

 

Question 17. Evaluate the Given limit: \( \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} \)
Answer: When we insert \( x = 0 \), the mathematical function produces the indeterminate form \( \frac{0}{0} \).
By using double-angle and half-angle formulas to change cosine to sine terms, we simplify:
\( \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} = \lim_{x \to 0} \frac{(1 - 2\sin^2 x) - 1}{\left(1 - 2\sin^2 \frac{x}{2}\right) - 1} \) [using \( \cos \theta = 1 - 2\sin^2 \frac{\theta}{2} \)]
\( = \lim_{x \to 0} \frac{-2\sin^2 x}{-2\sin^2 \frac{x}{2}} \)
\( = \lim_{x \to 0} \frac{\sin^2 x}{\sin^2 \frac{x}{2}} \)
\( = \lim_{x \to 0} \left[ \frac{\left( \frac{\sin^2 x}{x^2} \right) \times x^2}{\left( \frac{\sin^2 \frac{x}{2}}{\left( \frac{x}{2} \right)^2} \right) \times \frac{x^2}{4}} \right] \)
\( = 4 \times \frac{\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2}{\lim_{\frac{x}{2} \to 0} \left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)^2} \) [since \( x \to 0 \implies \frac{x}{2} \to 0 \)]
\( = 4 \times \frac{1^2}{1^2} \) [using standard limit \( \lim_{y \to 0} \frac{\sin y}{y} = 1 \)]
\( = 4 \)
In simple words: Replace the cosines using trigonometric identity formulas to convert them into sines. Then divide by \( x^2 \) to match standard limit templates, which helps us simplify to 4.

Exam Tip: The identity \( 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \) is an exceptionally common tool used to eliminate \( \frac{0}{0} \) forms in limit problems.

 

Question 18. Evaluate the Given limit: \( \lim_{x \to 0} \frac{ax + x\cos x}{b\sin x} \)
Answer: If \( x = 0 \) is substituted, the expression outputs the indeterminate form \( \frac{0}{0} \).
Factoring out the term \( x \) from the top part, we get:
\( \lim_{x \to 0} \frac{ax + x\cos x}{b\sin x} = \frac{1}{b} \lim_{x \to 0} \frac{x(a + \cos x)}{\sin x} \)
\( = \frac{1}{b} \lim_{x \to 0} \left( \frac{x}{\sin x} \right) \times \lim_{x \to 0} (a + \cos x) \)
\( = \frac{1}{b} \times \frac{1}{\lim_{x \to 0} \frac{\sin x}{x}} \times \lim_{x \to 0} (a + \cos x) \)
\( = \frac{1}{b} \times \frac{1}{1} \times (a + \cos 0) \)
\( = \frac{a + 1}{b} \)
In simple words: Pull out x from the top. We can rewrite the expression using \( \frac{x}{\sin x} \) which is just 1 when x goes to 0. This lets us solve the remaining part by putting in 0, giving us \( \frac{a+1}{b} \).

Exam Tip: Dividing or multiplying by variable groupings to isolate the \( \frac{\sin x}{x} \) expression is standard practice for clearing trigonometric indeterminacy.

 

Question 19. Evaluate the Given limit: \( \lim_{x \to 0} x\sec x \)
Answer: By using the reciprocal trigonometric relationship, we rewrite:
\( \lim_{x \to 0} x\sec x = \lim_{x \to 0} \frac{x}{\cos x} \)
\( = \frac{0}{\cos 0} \)
\( = \frac{0}{1} \)
\( = 0 \)
In simple words: Write secant as 1 over cosine. Plugging in 0 gives 0 divided by 1, which simply equals 0.

Exam Tip: Always convert secant, cosecant, cotangent, and tangent functions into sine and cosine terms to evaluate limit forms clearly.

 

Question 20. Evaluate the Given limit: \( \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} , a, b, a+b \neq 0 \)
Answer: Evaluating the expression at \( x = 0 \) leads to the indeterminate form \( \frac{0}{0} \).
To resolve this, divide both the numerator and denominator by \( x \):
\( \lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} = \lim_{x \to 0} \frac{\left( \frac{\sin ax}{x} \right) + b}{a + \left( \frac{\sin bx}{x} \right)} \)
\( = \lim_{x \to 0} \frac{a \left( \frac{\sin ax}{ax} \right) + b}{a + b \left( \frac{\sin bx}{bx} \right)} \)
\( = \frac{a \left( \lim_{ax \to 0} \frac{\sin ax}{ax} \right) + b}{a + b \left( \lim_{bx \to 0} \frac{\sin bx}{bx} \right)} \) [since \( x \to 0 \implies ax \to 0 \) and \( bx \to 0 \)]
\( = \frac{a(1) + b}{a + b(1)} \)
\( = \frac{a + b}{a + b} \)
\( = 1 \) [since \( a + b \neq 0 \)]
In simple words: Divide every term on top and bottom by x. This allows us to apply the standard rule to make the sines evaluate to 1. The fraction then simplifies to \( \frac{a+b}{a+b} \), which is 1.

Exam Tip: Be mindful of algebraic statements such as \( a + b \neq 0 \); they mathematically justify canceling the final terms in fractional outputs.

 

Question 21. Evaluate the Given limit: \( \lim_{x \to 0} (\csc x - \cot x) \)
Answer: Substituting \( x = 0 \) directly gives the indeterminate difference \( \infty - \infty \).
By rewriting the terms using sine and cosine functions, we obtain:
\( \lim_{x \to 0} (\csc x - \cot x) = \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) \)
\( = \lim_{x \to 0} \left( \frac{1 - \cos x}{\sin x} \right) \)
\( = \lim_{x \to 0} \left[ \frac{\frac{1 - \cos x}{x}}{\frac{\sin x}{x}} \right] \)
\( = \frac{\lim_{x \to 0} \left( \frac{1 - \cos x}{x} \right)}{\lim_{x \to 0} \left( \frac{\sin x}{x} \right)} \)
\( = \frac{0}{1} \)
\( = 0 \)
In simple words: Change cosecant to 1/sin and cotangent to cos/sin. Merge them into one fraction, and divide the top and bottom by x to use standard limits, resulting in 0 over 1, which equals 0.

Exam Tip: The identity \( \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \) is a standard limit that can be applied directly to simplify complex fractions.

 

Question 22. Evaluate the Given limit: \( \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}} \)
Answer: Evaluating at \( x = \frac{\pi}{2} \) gives the indeterminate form \( \frac{0}{0} \).
Let us substitute \( x - \frac{\pi}{2} = y \), meaning that \( y \to 0 \) as \( x \to \frac{\pi}{2} \).
This gives us \( x = y + \frac{\pi}{2} \).
Consequently,
\( \lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x - \frac{\pi}{2}} = \lim_{y \to 0} \frac{\tan 2\left( y + \frac{\pi}{2} \right)}{y} \)
\( = \lim_{y \to 0} \frac{\tan(\pi + 2y)}{y} \)
\( = \lim_{y \to 0} \frac{\tan 2y}{y} \) [since \( \tan(\pi + \theta) = \tan \theta \)]
\( = \lim_{y \to 0} \frac{\sin 2y}{y \cos 2y} \)
\( = \lim_{y \to 0} \left( \frac{\sin 2y}{2y} \times \frac{2}{\cos 2y} \right) \)
\( = \left( \lim_{2y \to 0} \frac{\sin 2y}{2y} \right) \times \left( \lim_{y \to 0} \frac{2}{\cos 2y} \right) \) [since \( y \to 0 \implies 2y \to 0 \)]
\( = 1 \times \frac{2}{\cos 0} \)
\( = 1 \times \frac{2}{1} \)
\( = 2 \)
In simple words: Since we are not taking the limit at 0, we change the variable by setting \( y = x - \frac{\pi}{2} \) to move the limit to 0. This lets us simplify the tangent function and apply standard limits to find 2.

Exam Tip: When the limit variable does not approach zero, substituting \( y = x - c \) is a vital technique to realign the limit to zero and apply standard rules.

 

Question 23. Find \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 1} f(x) \), where \( f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases} \)
Answer: We are given the piecewise function:
\( f(x) = \begin{cases} 2x + 3, & x \le 0 \\ 3(x + 1), & x > 0 \end{cases} \)

To find \( \lim_{x \to 0} f(x) \):
Left-Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} [2x + 3] = 2(0) + 3 = 3 \)

Right-Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} [3(x + 1)] = 3(0 + 1) = 3 \)

Because the left-hand limit equals the right-hand limit, the limit exists:
\( \lim_{x \to 0} f(x) = 3 \)

To find \( \lim_{x \to 1} f(x) \):
Since the point \( x = 1 \) is strictly greater than 0, we only evaluate using the second sub-function \( 3(x + 1) \):
L.H.L.: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} [3(x + 1)] = 3(1 + 1) = 6 \)
R.H.L.: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} [3(x + 1)] = 3(1 + 1) = 6 \)
Since L.H.L. = R.H.L., we get:
\( \lim_{x \to 1} f(x) = 6 \)
In simple words: For the limit at 0, check the equations on both sides of 0. Since both approaches give 3, the limit is 3. For the limit at 1, we are fully inside the second formula, so we just substitute 1 to get 6.

Exam Tip: For piecewise functions, you must evaluate both Left-Hand and Right-Hand limits at the boundary point (where the function formula changes) to verify if the general limit exists.

 

Question 24. Find \( \lim_{x \to 1} f(x) \), where \( f(x) = \begin{cases} x^2 - 1, & x \le 1 \\ -x^2 - 1, & x > 1 \end{cases} \)
Answer: We are given the piecewise function:
\( f(x) = \begin{cases} x^2 - 1, & x \le 1 \\ -x^2 - 1, & x > 1 \end{cases} \)

We evaluate the limit at the boundary point \( x = 1 \):
L.H.L.: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} [x^2 - 1] = 1^2 - 1 = 0 \)
R.H.L.: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} [-x^2 - 1] = -1^2 - 1 = -1 - 1 = -2 \)

We can easily see that the Left-Hand Limit does not equal the Right-Hand Limit:
\( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \)

Hence, \( \lim_{x \to 1} f(x) \) does not exist.
In simple words: Approaching 1 from the left gives 0, but approaching 1 from the right gives -2. Since the two sides do not meet at the same value, the general limit does not exist.

Exam Tip: If the L.H.L and R.H.L are unequal at a given point, write a clear concluding statement explaining why the limit fails to exist.

 

Question 25. Evaluate \( \lim_{x \to 0} f(x) \), where \( f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \)
Answer: We are given the function:
\( f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \)

We check the limit as \( x \) approaches 0 from both sides:
L.H.L.: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x} \)
Since \( x < 0 \implies |x| = -x \), we have:
\( \lim_{x \to 0^-} \left( \frac{-x}{x} \right) = \lim_{x \to 0^-} (-1) = -1 \)

R.H.L.: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x} \)
Since \( x > 0 \implies |x| = x \), we have:
\( \lim_{x \to 0^+} \left( \frac{x}{x} \right) = \lim_{x \to 0^+} (1) = 1 \)

Since L.H.L. \( \neq \) R.H.L., the overall limit \( \lim_{x \to 0} f(x) \) does not exist.
In simple words: When x is slightly negative, \( |x|/x \) simplifies to -1. When x is slightly positive, \( |x|/x \) simplifies to 1. Since the two sides approach different numbers, there is no limit.

Exam Tip: For any absolute value expression \( |x| \) near zero, always split the evaluation into \( x > 0 \) and \( x < 0 \) to determine the correct sign.

 

Question 26. Find \( \lim_{x \to 0} f(x) \), where \( f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases} \)
Answer: We are given the function:
\( f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases} \)

Let us analyze the limits from left and right:
L.H.L.: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} \)
Since \( x < 0 \implies |x| = -x \), we get:
\( \lim_{x \to 0^-} \left( \frac{x}{-x} \right) = \lim_{x \to 0^-} (-1) = -1 \)

R.H.L.: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|} \)
Since \( x > 0 \implies |x| = x \), we get:
\( \lim_{x \to 0^+} \left( \frac{x}{x} \right) = \lim_{x \to 0^+} (1) = 1 \)

As the Left-Hand Limit is not equal to the Right-Hand Limit, \( \lim_{x \to 0} f(x) \) does not exist.
In simple words: This function is similar to the previous one. From the left side of zero, the ratio is always -1, and from the right side, it is always 1. Because these values do not match, the limit cannot exist.

Exam Tip: The functions \( \frac{x}{|x|} \) and \( \frac{|x|}{x} \) represent the Signum function (excluding \( x=0 \)), which consistently has a jump discontinuity at the origin.

 

Question 27. Find \( \lim_{x \to 5} f(x) \), where \( f(x) = |x| - 5 \)
Answer: We are given the absolute value function \( f(x) = |x| - 5 \).
Because we are checking the limit near 5, x is strictly positive, hence we can replace \( |x| \) with \( x \):
L.H.L.: \( \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (|x| - 5) = \lim_{x \to 5^-} (x - 5) = 5 - 5 = 0 \)
R.H.L.: \( \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (|x| - 5) = \lim_{x \to 5^+} (x - 5) = 5 - 5 = 0 \)

Since L.H.L. = R.H.L. = 0, we conclude:
\( \lim_{x \to 5} f(x) = 0 \)
In simple words: Near 5, x is a positive number, so \( |x| \) is simply x. The function behaves like \( x - 5 \). Substituting 5 into this expression gives 0 from both sides.

Exam Tip: Since \( x = 5 \) is far from the turning point of the absolute value function (which is \( x = 0 \)), \( |x| \) is smoothly continuous and can be replaced directly with \( x \).

 

Question 28. Suppose \( f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases} \) and if \( \lim_{x \to 1} f(x) = f(1) \) what are possible values of \( a \) and \( b \)?
Answer: We are given the piecewise function:
\( f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases} \)

We first find the left and right limits at \( x = 1 \):
L.H.L.: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a + bx) = a + b(1) = a + b \)
R.H.L.: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (b - ax) = b - a(1) = b - a \)

We are given that \( \lim_{x \to 1} f(x) = f(1) \). This implies that:
L.H.L. = R.H.L. = \( f(1) \)

Using \( f(1) = 4 \), we set up the system of equations:
\( a + b = 4 \) --- (Equation 1)
\( b - a = 4 \implies -a + b = 4 \) --- (Equation 2)

By adding Equation 1 and Equation 2, we obtain:
\( 2b = 8 \implies b = 4 \)

Substituting \( b = 4 \) back into Equation 1:
\( a + 4 = 4 \implies a = 0 \)

Thus, the possible values are \( a = 0 \) and \( b = 4 \).
In simple words: Since the limit equals the function's value of 4 at \( x = 1 \), both the left equation \( a + b \) and the right equation \( b - a \) must equal 4. Solving these equations together gives \( a = 0 \) and \( b = 4 \).

Exam Tip: The condition \( \lim_{x \to c} f(x) = f(c) \) is the formal definition of continuity at a point; use it to set up simultaneous equations for unknown constants.

 

Question 29. Let \( a_1, a_2, \dots, a_n \) be fixed real numbers and define a function \( f(x) = (x - a_1)(x - a_2)\dots(x - a_n) \). What is \( \lim_{x \to a_1} f(x) \)? For some \( a \neq a_1, a_2, \dots, a_n \), compute \( \lim_{x \to a} f(x) \).
Answer: We are provided with the function:
\( f(x) = (x - a_1)(x - a_2)\dots(x - a_n) \)

To find the limit as \( x \) approaches \( a_1 \):
\( \lim_{x \to a_1} f(x) = \lim_{x \to a_1} [(x - a_1)(x - a_2)\dots(x - a_n)] \)
\( = \left[ \lim_{x \to a_1} (x - a_1) \right] \left[ \lim_{x \to a_1} (x - a_2) \right] \dots \left[ \lim_{x \to a_1} (x - a_n) \right] \)
\( = (a_1 - a_1)(a_1 - a_2)\dots(a_1 - a_n) \)
\( = 0 \times (a_1 - a_2)\dots(a_1 - a_n) \)
\( = 0 \)

Next, we evaluate the limit as \( x \) approaches a generic point \( a \) (where \( a \neq a_i \)):
\( \lim_{x \to a} f(x) = \lim_{x \to a} [(x - a_1)(x - a_2)\dots(x - a_n)] \)
\( = \left[ \lim_{x \to a} (x - a_1) \right] \left[ \lim_{x \to a} (x - a_2) \right] \dots \left[ \lim_{x \to a} (x - a_n) \right] \)
\( = (a - a_1)(a - a_2)\dots(a - a_n) \)
In simple words: Substituting \( a_1 \) in place of x makes the very first bracket 0. Since we are multiplying all brackets together, multiplying by 0 makes the whole product 0. For any other number a, simply swap x with a to get the product of those subtracted terms.

Exam Tip: Since polynomial functions are continuous everywhere, you can always determine their limits by directly substituting the limiting value into the variable.

 

Question 30. If \( f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases} \). For what value (s) of a does \( \lim_{x \to a} f(x) \) exists?
Answer: We are given the piecewise function:
\( f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases} \)

To solve this, we must examine the limit at three distinct scenarios for the constant a:

Case I: When \( a = 0 \)
Left-Hand Limit (L.H.L.):
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (|x| + 1) \)
Since \( x < 0 \implies |x| = -x \), we have:
\( \lim_{x \to 0^-} (-x + 1) = -0 + 1 = 1 \)

Right-Hand Limit (R.H.L.):
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (|x| - 1) \)
Since \( x > 0 \implies |x| = x \), we have:
\( \lim_{x \to 0^+} (x - 1) = 0 - 1 = -1 \)

Since L.H.L. \( \neq \) R.H.L. at \( x = 0 \), the limit \( \lim_{x \to 0} f(x) \) does not exist.

Case II: When \( a < 0 \)
L.H.L.: \( \lim_{x \to a^-} f(x) = \lim_{x \to a^-} (|x| + 1) \)
Since \( a < 0 \implies |x| = -x \):
\( = \lim_{x \to a^-} (-x + 1) = -a + 1 \)

R.H.L.: \( \lim_{x \to a^+} f(x) = \lim_{x \to a^+} (|x| + 1) \)
Since \( a < 0 \implies |x| = -x \):
\( = \lim_{x \to a^+} (-x + 1) = -a + 1 \)
Since L.H.L. = R.H.L. = \( -a + 1 \), the limit exists for all values \( a < 0 \).

Case III: When \( a > 0 \)
L.H.L.: \( \lim_{x \to a^-} f(x) = \lim_{x \to a^-} (|x| - 1) \)
Since \( a > 0 \implies |x| = x \):
\( = \lim_{x \to a^-} (x - 1) = a - 1 \)

R.H.L.: \( \lim_{x \to a^+} f(x) = \lim_{x \to a^+} (|x| - 1) \)
Since \( a > 0 \implies |x| = x \):
\( = \lim_{x \to a^+} (x - 1) = a - 1 \)
Since L.H.L. = R.H.L. = \( a - 1 \), the limit exists for all values \( a > 0 \).

Thus, the limit \( \lim_{x \to a} f(x) \) exists for all real numbers \( a \neq 0 \).
In simple words: At 0, the left approach goes to 1 but the right approach goes to -1, so no limit exists there. At any positive or negative number, the function is smoothly defined on both sides, so the limit exists for any value of a as long as a is not 0.

Exam Tip: When checking limits of absolute values at points other than zero, state whether \( x \) is positive or negative so you can drop the absolute value symbol mathematically.

 

Question 31. If the function \( f(x) \) satisfies \( \lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi \), evaluate \( \lim_{x \to 1} f(x) \).
Answer: We start with the given equation:
\( \lim_{x \to 1} \frac{f(x) - 2}{x^2 - 1} = \pi \)

By applying basic limit quotient rules, we write:
\( \frac{\lim_{x \to 1} [f(x) - 2]}{\lim_{x \to 1} (x^2 - 1)} = \pi \)

Multiplying the denominator limit to both sides yields:
\( \lim_{x \to 1} [f(x) - 2] = \pi \times \lim_{x \to 1} (x^2 - 1) \)
\( \lim_{x \to 1} [f(x) - 2] = \pi \times (1^2 - 1) \)
\( \lim_{x \to 1} [f(x) - 2] = \pi \times 0 \)
\( \lim_{x \to 1} [f(x) - 2] = 0 \)

Using the difference rule for limits:
\( \lim_{x \to 1} f(x) - \lim_{x \to 1} 2 = 0 \)
\( \lim_{x \to 1} f(x) - 2 = 0 \)
\( \lim_{x \to 1} f(x) = 2 \)
In simple words: Since the denominator goes to 0 as x goes to 1, the numerator must also go to 0 for the fraction to have a finite limit of \( \pi \). Setting the limit of the top part to 0 tells us that the limit of f(x) must equal 2.

Exam Tip: For any limit of the form \( \lim \frac{g(x)}{h(x)} = L \), if \( \lim h(x) = 0 \) and \( L \) is a finite real number, then \( \lim g(x) \) must always equal 0.

 

Question 32. If \( f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3 + m, & x > 1 \end{cases} \). For what integers \( m \) and \( n \) does \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 1} f(x) \) exist?
Answer: The function is defined as:
\( f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3 + m, & x > 1 \end{cases} \)

First, we analyze \( \lim_{x \to 0} f(x) \):
L.H.L.: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (mx^2 + n) = m(0)^2 + n = n \)
R.H.L.: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (nx + m) = n(0) + m = m \)
Consequently, this limit exists only when the left and right limits match:
\( m = n \)

Next, we analyze \( \lim_{x \to 1} f(x) \):
L.H.L.: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (nx + m) = n(1) + m = m + n \)
R.H.L.: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (nx^3 + m) = n(1)^3 + m = m + n \)
Since L.H.L. = R.H.L. = \( m + n \), this limit exists for any integral values of \( m \) and \( n \).

In summary, \( \lim_{x \to 0} f(x) \) exists only if \( m = n \), while \( \lim_{x \to 1} f(x) \) exists for all integer values of \( m \) and \( n \).
In simple words: At 0, the left limit is n and the right limit is m, so they must be equal for the limit to exist. At 1, both the left and right equations evaluate to m + n, which means they are always equal regardless of what integers you choose for m and n.

Exam Tip: Be sure to write separate, clear concluding sentences for each of the point limits evaluated when a question asks for conditions at multiple boundaries.

Exercise 13.2

 

Question 1. Find the derivative of \( x^2 - 2 \) at \( x = 10 \).
Answer: Let the function be \( f(x) = x^2 - 2 \). Therefore,
\( f'(10) = \lim_{h \to 0} \frac{f(10 + h) - f(10)}{h} \)
\( = \lim_{h \to 0} \frac{[(10 + h)^2 - 2] - (10^2 - 2)}{h} \)
\( = \lim_{h \to 0} \frac{10^2 + 2(10)h + h^2 - 2 - 10^2 + 2}{h} \)
\( = \lim_{h \to 0} \frac{20h + h^2}{h} \)
\( = \lim_{h \to 0} (20 + h) \)
\( = 20 + 0 \)
\( = 20 \)

Thus, the derivative of the expression at \( x = 10 \) is 20.
In simple words: To find the rate of change at 10, we use the derivative formula with a small gap h. We plug in 10 + h and 10, expand the squares, cancel matching terms, divide by h, and let h become 0 to get 20.

Exam Tip: Always show the limit expansion steps clearly when asked to find the derivative at a point before performing calculations.

 

Question 2. Find the derivative of \( 99x \) at \( x = 100 \).
Answer: Let us define \( f(x) = 99x \). Thus,
\( f'(100) = \lim_{h \to 0} \frac{f(100 + h) - f(100)}{h} \)
\( = \lim_{h \to 0} \frac{99(100 + h) - 99(100)}{h} \)
\( = \lim_{h \to 0} \frac{99(100) + 99h - 99(100)}{h} \)
\( = \lim_{h \to 0} \frac{99h}{h} \)
\( = \lim_{h \to 0} (99) \)
\( = 99 \)

Thus, the derivative of \( 99x \) at \( x = 100 \) is 99.
In simple words: The derivative of a linear function like 99x is just its constant slope, which is 99. Using the formula, the h factors cancel out directly, leaving us with 99.

Exam Tip: Remember that the derivative of any linear function \( f(x) = mx + c \) is always its slope \( m \), regardless of the value of \( x \).

 

Question 3. Find the derivative of \( x \) at \( x = 1 \).
Answer: Let \( f(x) = x \). Consequently,
\( f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \)
\( = \lim_{h \to 0} \frac{(1 + h) - 1}{h} \)
\( = \lim_{h \to 0} \frac{h}{h} \)
\( = \lim_{h \to 0} (1) \)
\( = 1 \)

Thus, the derivative of \( x \) at \( x = 1 \) is 1.
In simple words: The derivative of x is simply 1. We compute this by dividing the change in the function, which is h, by h, giving us a constant value of 1.

Exam Tip: Treat the function \( f(x) = x \) as a special case of \( x^n \) where \( n = 1 \); its derivative is always 1 for all real numbers.

 

Question 4. Find the derivative of the following functions from first principle.
(i) \( x^3 - 27 \)
(ii) \( (x - 1)(x - 2) \)
(iii) \( \frac{1}{x^2} \)
(iv) \( \frac{x+1}{x-1} \)
Answer:
(i) Let \( f(x) = x^3 - 27 \). Applying the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{[(x + h)^3 - 27] - (x^3 - 27)}{h} \)
\( = \lim_{h \to 0} \frac{(x^3 + h^3 + 3x^2h + 3xh^2 - 27) - x^3 + 27}{h} \)
\( = \lim_{h \to 0} \frac{h^3 + 3x^2h + 3xh^2}{h} \)
\( = \lim_{h \to 0} (h^2 + 3x^2 + 3xh) \)
\( = 0 + 3x^2 + 0 \)
\( = 3x^2 \)

(ii) Let \( f(x) = (x - 1)(x - 2) \). Applying the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{[(x + h - 1)(x + h - 2)] - [(x - 1)(x - 2)]}{h} \)
\( = \lim_{h \to 0} \frac{(x^2 + hx - 2x + hx + h^2 - 2h - x - h + 2) - (x^2 - 3x + 2)}{h} \)
\( = \lim_{h \to 0} \frac{(2hx + h^2 - 3h)}{h} \)
\( = \lim_{h \to 0} (2x + h - 3) \)
\( = 2x + 0 - 3 \)
\( = 2x - 3 \)

(iii) Let \( f(x) = \frac{1}{x^2} \). Applying the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x^2 - (x+h)^2}{x^2(x+h)^2} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x^2 - (x^2 + h^2 + 2hx)}{x^2(x+h)^2} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-h^2 - 2hx}{x^2(x+h)^2} \right] \)
\( = \lim_{h \to 0} \left[ \frac{-h - 2x}{x^2(x+h)^2} \right] \)
\( = \frac{0 - 2x}{x^2(x+0)^2} \)
\( = -\frac{2x}{x^4} \)
\( = -\frac{2}{x^3} \)

(iv) Let \( f(x) = \frac{x+1}{x-1} \). Applying the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{(x^2 + hx + x - x - h - 1) - (x^2 + hx - x + x + h - 1)}{(x+h-1)(x-1)} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-2h}{(x+h-1)(x-1)} \right] \)
\( = \lim_{h \to 0} \left[ \frac{-2}{(x+h-1)(x-1)} \right] \)
\( = \frac{-2}{(x-1)(x-1)} \)
\( = -\frac{2}{(x-1)^2} \)
In simple words: The first principles method uses algebraic expansion, common denominators, or algebraic simplification to isolate the gap h in the numerator, cancel it with the h in the denominator, and then set h to 0.

Exam Tip: "First principle" explicitly means you must use the limit definition \( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) to derive the function; do not write down derivative rules directly without showing the limit steps.

 

Question 5. For the function \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1 \). Prove that \( f'(1) = 100 f'(0) \).
Answer: We are given the following polynomial function:
\( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1 \)

Taking the derivative of the function with respect to x:
\( f'(x) = \frac{d}{dx} \left[ \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1 \right] \)
\( = \frac{1}{100}(100x^{99}) + \frac{1}{99}(99x^{98}) + \dots + \frac{1}{2}(2x) + 1 + 0 \) [applying standard power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \)]
\( = x^{99} + x^{98} + \dots + x + 1 \)

Now, we evaluate the derivative at \( x = 0 \):
\( f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1 \)

Next, we evaluate the derivative at \( x = 1 \):
\( f'(1) = 1^{99} + 1^{98} + \dots + 1 + 1 \)
Since there are 100 terms in this sequence (from power 1 to power 99, plus the constant 1):
\( f'(1) = [1 + 1 + \dots + 1]_{100 \text{ terms}} = 100 \times 1 = 100 \)

Comparing the two evaluated values:
\( f'(1) = 100 \)
Since \( f'(0) = 1 \), we can write:
\( f'(1) = 100 \times f'(0) \)

This completes our proof.
In simple words: When we find the derivative of the function, the fraction coefficients cancel out nicely, leaving us with a sum of powers of x from 99 down to 1, plus 1. Plugging in 0 gives 1, while plugging in 1 adds up 100 ones, which equals 100. Thus, f'(1) is 100 times f'(0).

Exam Tip: Be careful when counting the terms in a polynomial series to ensure you don't miss the constant term when evaluating at \( x = 1 \).

 

Question 6. Find the derivative of \( x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n \) for some fixed real number \( a \).
Answer: Let \( f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n \).
Taking the derivative with respect to x:
\( f'(x) = \frac{d}{dx} (x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n) \)
\( = \frac{d}{dx}(x^n) + a\frac{d}{dx}(x^{n-1}) + a^2\frac{d}{dx}(x^{n-2}) + \dots + a^{n-1}\frac{d}{dx}(x) + a^n\frac{d}{dx}(1) \)

Applying the standard power rule formula:
\( f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1}(1) + a^n(0) \)
\( = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1} \)
In simple words: Differentiate each term of the series separately. Note that 'a' is a constant, so it remains as-is, and the last term \( a^n \) has no x, so its derivative is 0. All other terms follow the power rule \( nx^{n-1} \).

Exam Tip: Always state clearly that terms containing only constants (like \( a^n \)) have a derivative of zero with respect to the variable \( x \).

 

Question 7. For some constants \( a \) and \( b \), find the derivative of
(i) \( (x - a)(x - b) \)
(ii) \( (ax^2 + b)^2 \)
(iii) \( \frac{x-a}{x-b} \)
Answer:
(i) Let \( f(x) = (x - a)(x - b) \):
Expanding the brackets:
\( f(x) = x^2 - (a + b)x + ab \)
Differentiating with respect to x:
\( f'(x) = \frac{d}{dx} [x^2 - (a + b)x + ab] \)
\( = \frac{d}{dx}(x^2) - (a+b)\frac{d}{dx}(x) + \frac{d}{dx}(ab) \)
\( = 2x - (a+b)(1) + 0 \)
\( = 2x - a - b \)

(ii) Let \( f(x) = (ax^2 + b)^2 \):
Expanding using the algebraic identity \( (u + v)^2 \):
\( f(x) = a^2x^4 + 2abx^2 + b^2 \)
Differentiating with respect to x:
\( f'(x) = \frac{d}{dx} (a^2x^4 + 2abx^2 + b^2) \)
\( = a^2(4x^3) + 2ab(2x) + 0 \)
\( = 4a^2x^3 + 4abx \)
\( = 4ax(ax^2 + b) \)

(iii) Let \( f(x) = \frac{x-a}{x-b} \):
Applying the quotient rule of differentiation:
\( f'(x) = \frac{(x-b)\frac{d}{dx}(x-a) - (x-a)\frac{d}{dx}(x-b)}{(x-b)^2} \)
\( = \frac{(x-b)(1 - 0) - (x-a)(1 - 0)}{(x-b)^2} \)
\( = \frac{x - b - x + a}{(x-b)^2} \)
\( = \frac{a - b}{(x-b)^2} \)
In simple words: For parts (i) and (ii), expand the algebraic expressions first, then differentiate term-by-term using the power rule. For part (iii), use the quotient rule: (bottom times derivative of top minus top times derivative of bottom) divided by bottom squared.

Exam Tip: Expanding polynomial products before differentiating is often much faster and less error-prone than applying the product rule directly.

 

Question 8. Find the derivative of \( \frac{x^n - a^n}{x - a} \) for some constant \( a \).
Answer: Let the rational function be \( f(x) = \frac{x^n - a^n}{x - a} \). Using the quotient rule of derivatives:
\( f'(x) = \frac{(x-a)\frac{d}{dx}(x^n - a^n) - (x^n - a^n)\frac{d}{dx}(x-a)}{(x-a)^2} \)
\( = \frac{(x-a)(nx^{n-1} - 0) - (x^n - a^n)(1 - 0)}{(x-a)^2} \)
\( = \frac{(x-a)(nx^{n-1}) - (x^n - a^n)}{(x-a)^2} \)
\( = \frac{nx^n - anx^{n-1} - x^n + a^n}{(x-a)^2} \)
\( = \frac{x^n(n - 1) - anx^{n-1} + a^n}{(x-a)^2} \)
In simple words: Apply the quotient rule. The derivative of \( x^n - a^n \) is \( nx^{n-1} \) (since \( a^n \) is constant), and the derivative of \( x - a \) is 1. Multiply out the terms in the numerator to get the simplified final fraction.

Exam Tip: Remember that \( a \) is a constant, so any power of \( a \) (like \( a^n \)) has a derivative of zero. Do not mistake it for a variable and apply the power rule to it.

 

Question 9. Find the derivative of
(i) \( 2x - \frac{3}{4} \)
(ii) \( (5x^3 + 3x - 1)(x - 1) \)
(iii) \( x^{-3}(5 + 3x) \)
(iv) \( x^5(3 - 6x^{-9}) \)
(v) \( x^{-4}(3 - 4x^{-5}) \)
(vi) \( \frac{2}{x+1} - \frac{x^2}{3x-1} \)
Answer:
(i) Let \( f(x) = 2x - \frac{3}{4} \):
Taking the derivative with respect to x:
\( f'(x) = \frac{d}{dx} \left(2x - \frac{3}{4}\right) = 2(1) - 0 = 2 \)

(ii) Let \( f(x) = (5x^3 + 3x - 1)(x - 1) \):
Applying Leibnitz's product rule of differentiation:
\( f'(x) = (5x^3 + 3x - 1)\frac{d}{dx}(x-1) + (x-1)\frac{d}{dx}(5x^3 + 3x - 1) \)
\( = (5x^3 + 3x - 1)(1) + (x-1)(15x^2 + 3) \)
\( = 5x^3 + 3x - 1 + 15x^3 + 3x - 15x^2 - 3 \)
\( = 20x^3 - 15x^2 + 6x - 4 \)

(iii) Let \( f(x) = x^{-3}(5 + 3x) \):
Applying Leibnitz's product rule of differentiation:
\( f'(x) = x^{-3}\frac{d}{dx}(5+3x) + (5+3x)\frac{d}{dx}(x^{-3}) \)
\( = x^{-3}(3) + (5+3x)(-3x^{-4}) \)
\( = 3x^{-3} - 15x^{-4} - 9x^{-3} \)
\( = -6x^{-3} - 15x^{-4} \)
\( = -3x^{-3} \left( 2 + \frac{5}{x} \right) \)
\( = -\frac{3(5+2x)}{x^4} \)

(iv) Let \( f(x) = x^5(3 - 6x^{-9}) \):
Applying Leibnitz's product rule of differentiation:
\( f'(x) = x^5\frac{d}{dx}(3 - 6x^{-9}) + (3 - 6x^{-9})\frac{d}{dx}(x^5) \)
\( = x^5(0 - 6(-9)x^{-10}) + (3 - 6x^{-9})(5x^4) \)
\( = x^5(54x^{-10}) + 15x^4 - 30x^{-5} \)
\( = 54x^{-5} + 15x^4 - 30x^{-5} \)
\( = 24x^{-5} + 15x^4 \)
\( = 15x^4 + \frac{24}{x^5} \)

(v) Let \( f(x) = x^{-4}(3 - 4x^{-5}) \):
Applying Leibnitz's product rule of differentiation:
\( f'(x) = x^{-4}\frac{d}{dx}(3 - 4x^{-5}) + (3 - 4x^{-5})\frac{d}{dx}(x^{-4}) \)
\( = x^{-4}(20x^{-6}) + (3 - 4x^{-5})(-4x^{-5}) \)
\( = 20x^{-10} - 12x^{-5} + 16x^{-10} \)
\( = 36x^{-10} - 12x^{-5} \)
\( = -\frac{12}{x^5} + \frac{36}{x^{10}} \)

(vi) Let \( f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1} \):
Differentiating term-by-term using the quotient rule:
\( f'(x) = \frac{d}{dx}\left(\frac{2}{x+1}\right) - \frac{d}{dx}\left(\frac{x^2}{3x-1}\right) \)

For the first term:
\( \frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{(x+1)(0) - 2(1)}{(x+1)^2} = -\frac{2}{(x+1)^2} \)

For the second term:
\( \frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{(3x-1)(2x) - x^2(3)}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2} \)

Combining both parts:
\( f'(x) = -\frac{2}{(x+1)^2} - \frac{x(3x-2)}{(3x-1)^2} \)
In simple words: This problem covers product rule and quotient rule. For products, differentiate one part while keeping the other constant, and swap. For quotients, use the quotient formula on each term to arrive at the derivative.

Exam Tip: Be very careful when expanding negative powers; remember that multiplying identical base variables adds their exponents (e.g., \( x^{-4} \times x^{-6} = x^{-10} \)).

 

Question 10. Find the derivative of cos x from first principle.
Answer: We define \( f(x) = \cos x \). Applying the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \)
\( = \lim_{h \to 0} \left[ \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \right] \) [using compound angle formula for \( \cos(A+B) \)]
\( = \lim_{h \to 0} \left[ \frac{-\cos x(1 - \cos h) - \sin x \sin h}{h} \right] \)
\( = \lim_{h \to 0} \left[ -\cos x \left(\frac{1 - \cos h}{h}\right) - \sin x \left(\frac{\sin h}{h}\right) \right] \)
\( = -\cos x \left( \lim_{h \to 0} \frac{1 - \cos h}{h} \right) - \sin x \left( \lim_{h \to 0} \frac{\sin h}{h} \right) \)
Using the standard limits \( \lim_{h \to 0} \frac{1 - \cos h}{h} = 0 \) and \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \):
\( = -\cos x(0) - \sin x(1) \)
\( = -\sin x \)

Hence, \( f'(x) = -\sin x \).
In simple words: Expand \( \cos(x+h) \) using trigonometric formulas, group the \( \cos x \) terms, and separate the limits. The term with \( (1-\cos h)/h \) becomes 0, and \( \sin h/h \) becomes 1, leaving us with \( -\sin x \).

Exam Tip: Remember to write down standard limits used, such as \( \lim_{h\to 0} \frac{1-\cos h}{h} = 0 \) and \( \lim_{h\to 0} \frac{\sin h}{h} = 1 \), to secure full marks in first-principle questions.

 

Question 11. Find the derivative of the following functions:
(i) \( \sin x \cos x \)
(ii) \( \sec x \)
(iii) \( 5 \sec x + 4 \cos x \)
(iv) \( \csc x \)
(v) \( 3\cot x + 5\csc x \)
(vi) \( 5\sin x - 6\cos x + 7 \)
(vii) \( 2\tan x - 7\sec x \)
Answer:
(i) Let \( f(x) = \sin x \cos x \). Applying first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\sin(x+h)\cos(x+h) - \sin x \cos x}{h} \)
\( = \lim_{h \to 0} \frac{1}{2h} [2\sin(x+h)\cos(x+h) - 2\sin x \cos x] \)
\( = \lim_{h \to 0} \frac{\sin(2x+2h) - \sin 2x}{2h} \) [using identity \( 2\sin\theta\cos\theta = \sin 2\theta \)]
\( = \lim_{h \to 0} \frac{1}{2h} \left[ 2 \cos\left(\frac{2x+2h+2x}{2}\right) \sin\left(\frac{2x+2h-2x}{2}\right) \right] \) [using \( \sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{1}{2h} [2 \cos(2x+h) \sin h] \)
\( = \lim_{h \to 0} \cos(2x+h) \cdot \lim_{h \to 0} \frac{\sin h}{h} \)
\( = \cos(2x + 0) \cdot (1) \)
\( = \cos 2x \)

(ii) Let \( f(x) = \sec x \). Applying first principles:
\( f'(x) = \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{\cos(x+h)} - \frac{1}{\cos x} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\cos x - \cos(x+h)}{\cos x \cos(x+h)} \right] \)
\( = \lim_{h \to 0} \frac{1}{h \cos x \cos(x+h)} \left[ -2 \sin\left(\frac{x+x+h}{2}\right) \sin\left(\frac{x - x - h}{2}\right) \right] \) [using \( \cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{1}{h \cos x \cos(x+h)} \left[ 2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right) \right] \)
\( = \frac{1}{\cos x} \lim_{h \to 0} \left[ \frac{\sin\left(x+\frac{h}{2}\right)}{\cos(x+h)} \right] \cdot \lim_{h \to 0} \left[ \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = \frac{1}{\cos x} \left[ \frac{\sin x}{\cos x} \right] \cdot (1) \)
\( = \sec x \tan x \)

(iii) Let \( f(x) = 5\sec x + 4\cos x \). Applying first principles:
\( f'(x) = \lim_{h \to 0} \frac{[5\sec(x+h) + 4\cos(x+h)] - [5\sec x + 4\cos x]}{h} \)
\( = 5 \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h} + 4 \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \)
By utilizing the derivative results from earlier sections for \( \sec x \) and \( \cos x \):
\( = 5(\sec x \tan x) + 4(-\sin x) \)
\( = 5\sec x \tan x - 4\sin x \)
In simple words: This question solves standard trigonometric functions from first principles. By converting functions to sines and cosines, applying trigonometric sum-to-product identities, and evaluating the standard limits, we derive the exact rates of change.

Exam Tip: For composite trigonometric expressions like \( 5\sec x + 4\cos x \), you can split the limit into separate parts and apply individual derivative proofs to keep calculations clean and manageable.

 

Question 11. (Continued) Find the derivative of the following functions:
(iv) \( \csc x \)
(v) \( 3\cot x + 5\csc x \)
(vi) \( 5\sin x - 6\cos x + 7 \)
(vii) \( 2\tan x - 7\sec x \)
Answer:
(iv) Let \( f(x) = \csc x \). Evaluating using the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{\csc(x+h) - \csc x}{h} \)
\( = \lim_{h \to 0} \frac{\frac{1}{\sin(x+h)} - \frac{1}{\sin x}}{h} \)
\( = \lim_{h \to 0} \frac{\sin x - \sin(x+h)}{h \sin x \sin(x+h)} \)
\( = \lim_{h \to 0} \frac{2\cos\left(\frac{x+x+h}{2}\right)\sin\left(\frac{x-x-h}{2}\right)}{h \sin x \sin(x+h)} \) [using \( \sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{-2\cos\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h \sin x \sin(x+h)} \) [since \( \sin(-\theta) = -\sin\theta \)]
\( = -\frac{1}{\sin x} \lim_{h \to 0} \left[ \frac{\cos\left(x+\frac{h}{2}\right)}{\sin(x+h)} \right] \cdot \lim_{h \to 0} \left[ \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = -\frac{1}{\sin x} \left[ \frac{\cos x}{\sin x} \right] \cdot (1) \)
\( = -\csc x \cot x \)

(v) Let \( f(x) = 3\cot x + 5\csc x \). Evaluating using the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{[3\cot(x+h) + 5\csc(x+h)] - [3\cot x + 5\csc x]}{h} \)
\( = 3 \lim_{h \to 0} \frac{\cot(x+h) - \cot x}{h} + 5 \lim_{h \to 0} \frac{\csc(x+h) - \csc x}{h} \) --- (Equation 1)

First, we resolve the limit of the cotangent term:
\( \lim_{h \to 0} \frac{\cot(x+h) - \cot x}{h} = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\cos(x+h)}{\sin(x+h)} - \frac{\cos x}{\sin x} \right] \)
\( = \lim_{h \to 0} \frac{\cos(x+h)\sin x - \cos x \sin(x+h)}{h \sin x \sin(x+h)} \)
\( = \lim_{h \to 0} \frac{\sin(x - (x+h))}{h \sin x \sin(x+h)} \) [using the compound angle rule \( \sin(A-B) = \sin A\cos B - \cos A\sin B \)]
\( = \lim_{h \to 0} \frac{\sin(-h)}{h \sin x \sin(x+h)} \)
\( = -\frac{1}{\sin x} \left( \lim_{h \to 0} \frac{\sin h}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{\sin(x+h)} \right) \)
\( = -1 \cdot \frac{1}{\sin x \cdot \sin x} = -\csc^2 x \) --- (Equation 2)

From part (iv), we have established that:
\( \lim_{h \to 0} \frac{\csc(x+h) - \csc x}{h} = -\csc x \cot x \) --- (Equation 3)

Substituting Equation 2 and Equation 3 into Equation 1, we get:
\( f'(x) = -3\csc^2 x - 5\csc x \cot x \)

(vi) Let \( f(x) = 5\sin x - 6\cos x + 7 \). Evaluating using the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{[5\sin(x+h) - 6\cos(x+h) + 7] - [5\sin x - 6\cos x + 7]}{h} \)
\( = 5 \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} - 6 \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} + \lim_{h \to 0} \frac{7 - 7}{h} \)

By bringing together these distinct parts, we establish:
\( f'(x) = 5\cos x - 6(-\sin x) + 0 \)
\( = 5\cos x + 6\sin x \)

(vii) Let \( f(x) = 2\tan x - 7\sec x \). Evaluating using the definition of first principles:
\( f'(x) = \lim_{h \to 0} \frac{[2\tan(x+h) - 7\sec(x+h)] - [2\tan x - 7\sec x]}{h} \)
\( = 2 \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h} - 7 \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h} \)

First, we resolve the limit of the tangent term:
\( \lim_{h \to 0} \frac{\tan(x+h) - \tan x}{h} = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x} \right] \)
\( = \lim_{h \to 0} \frac{\sin(x+h)\cos x - \sin x\cos(x+h)}{h \cos x\cos(x+h)} \)
\( = \lim_{h \to 0} \frac{\sin(x+h-x)}{h \cos x\cos(x+h)} \) [using \( \sin(A-B) = \sin A\cos B - \cos A\sin B \)]
\( = \frac{1}{\cos x} \left( \lim_{h \to 0} \frac{\sin h}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{\cos(x+h)} \right) \)
\( = \frac{1}{\cos x} \cdot 1 \cdot \frac{1}{\cos x} = \sec^2 x \)

Since we proved earlier that the derivative of \( \sec x \) is \( \sec x \tan x \):
\( f'(x) = 2\sec^2 x - 7\sec x \tan x \)
In simple words: First principles for cosecant and cotangent functions are solved by replacing them with sines and cosines. Then, simplify the fractional fractions on top using common denominators, apply compound angle identities, and calculate the limit to find the derivatives.

Exam Tip: Keep basic trigonometric identities, especially compound angle formulas and sum-to-product transforms, memorized as they are essential for first principle proofs of trigonometric functions.

NCERT Miscellaneous Solutions

 

Question 1. Find the derivative of the following functions from first principle:
(i) \( -x \)
(ii) \( (-x)^{-1} \)
(iii) \( \sin(x+1) \)
(iv) \( \cos\left(x - \frac{\pi}{8}\right) \)
Answer:
(i) Let \( f(x) = -x \). This gives \( f(x+h) = -(x+h) \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} \)
\( = \lim_{h \to 0} \frac{-x - h + x}{h} \)
\( = \lim_{h \to 0} \frac{-h}{h} \)
\( = \lim_{h \to 0} (-1) = -1 \)

(ii) Let \( f(x) = (-x)^{-1} = -\frac{1}{x} \). This gives \( f(x+h) = -\frac{1}{x+h} \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{-\frac{1}{x+h} - \left(-\frac{1}{x}\right)}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ -\frac{1}{x+h} + \frac{1}{x} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-x + (x+h)}{x(x+h)} \right] \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{h}{x(x+h)} \right] \)
\( = \lim_{h \to 0} \frac{1}{x(x+h)} \)
\( = \frac{1}{x(x+0)} = \frac{1}{x^2} \)

(iii) Let \( f(x) = \sin(x+1) \). This gives \( f(x+h) = \sin(x+h+1) \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\sin(x+h+1) - \sin(x+1)}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ 2 \cos\left(\frac{x+h+1 + x+1}{2}\right) \sin\left(\frac{x+h+1 - (x+1)}{2}\right) \right] \) [using \( \sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{1}{h} \left[ 2 \cos\left(\frac{2x+h+2}{2}\right) \sin\left(\frac{h}{2}\right) \right] \)
\( = \lim_{h \to 0} \left[ \cos\left(x + \frac{h}{2} + 1\right) \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = \lim_{h \to 0} \cos\left(x + \frac{h}{2} + 1\right) \cdot \lim_{\frac{h}{2} \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \)
\( = \cos(x+1) \cdot (1) = \cos(x+1) \)

(iv) Let \( f(x) = \cos\left(x - \frac{\pi}{8}\right) \). This gives \( f(x+h) = \cos\left(x + h - \frac{\pi}{8}\right) \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{\cos\left(x+h-\frac{\pi}{8}\right) - \cos\left(x-\frac{\pi}{8}\right)}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ -2 \sin\left(\frac{x+h-\frac{\pi}{8} + x - \frac{\pi}{8}}{2}\right) \sin\left(\frac{x+h-\frac{\pi}{8} - \left(x-\frac{\pi}{8}\right)}{2}\right) \right] \) [using \( \cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{1}{h} \left[ -2 \sin\left(\frac{2x+h-\frac{\pi}{4}}{2}\right) \sin\left(\frac{h}{2}\right) \right] \)
\( = \lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2} - \frac{\pi}{8}\right) \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = -\sin\left(x - \frac{\pi}{8}\right) \cdot (1) = -\sin\left(x - \frac{\pi}{8}\right) \)
In simple words: First principles are calculated by inserting (x+h) into the equations, using common algebraic simplifications or trigonometric identities to isolate h, cancelling h, and then setting h to 0.

Exam Tip: Be mindful of signs when subtracting negative quantities like \( -x \) in part (i); writing parentheses clearly avoids sign mistakes in basic algebra steps.

 

Question 2. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (x + a) \)
Answer: Let us define \( f(x) = x + a \). This gives \( f(x+h) = x + h + a \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{(x+h+a) - (x+a)}{h} \)
\( = \lim_{h \to 0} \left(\frac{h}{h}\right) \)
\( = \lim_{h \to 0} (1) = 1 \)
In simple words: The derivative of \( x + a \) is simply 1. Since \( a \) is a constant, it has no effect on the rate of change of the expression.

Exam Tip: Since \( a \) is a constant, the slope of \( x + a \) is identical to the slope of \( x \), which is why the derivative evaluates simply to 1.

 

Question 3. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (px + q)\left(\frac{r}{x} + s\right) \)
Answer: Let \( f(x) = (px + q)\left(\frac{r}{x} + s\right) \).
Using Leibnitz's standard product rule:
\( f'(x) = (px+q)\frac{d}{dx}\left(\frac{r}{x}+s\right) + \left(\frac{r}{x}+s\right)\frac{d}{dx}(px+q) \)
\( = (px+q)\frac{d}{dx}(rx^{-1} + s) + \left(\frac{r}{x}+s\right)(p) \)
\( = (px+q)(-rx^{-2} + 0) + p\left(\frac{r}{x}+s\right) \)
\( = (px+q)\left(-\frac{r}{x^2}\right) + \frac{pr}{x} + ps \)
\( = -\frac{prx}{x^2} - \frac{qr}{x^2} + \frac{pr}{x} + ps \)
\( = -\frac{pr}{x} - \frac{qr}{x^2} + \frac{pr}{x} + ps \)
\( = ps - \frac{qr}{x^2} \)
In simple words: Apply the product rule. The derivative of the first term is p, and the derivative of the second term is \( -r/x^2 \). Multiply these results across the opposite brackets and simplify to find the final derivative.

Exam Tip: Expanding the brackets prior to taking the derivative is another safe way to solve this problem and verify your product rule solution.

 

Question 4. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (ax + b)(cx + d)^2 \)
Answer: Let \( f(x) = (ax+b)(cx+d)^2 \).
Using Leibnitz's standard product rule:
\( f'(x) = (ax+b)\frac{d}{dx}(cx+d)^2 + (cx+d)^2\frac{d}{dx}(ax+b) \)
\( = (ax+b)\frac{d}{dx}(c^2x^2 + 2cdx + d^2) + (cx+d)^2(a + 0) \)
\( = (ax+b)(2c^2x + 2cd) + a(cx+d)^2 \)
\( = 2c(ax+b)(cx+d) + a(cx+d)^2 \)
\( = (cx+d)[2c(ax+b) + a(cx+d)] \)
\( = (cx+d)[2acx + 2bc + acx + ad] \)
\( = (cx+d)[3acx + 2bc + ad] \)
In simple words: Differentiate the quadratic bracket and linear bracket using the product rule. Expand the first term using basic expansion rules, multiply, then factor out the common \( (cx + d) \) term to simplify the expression.

Exam Tip: Be sure to factor out common terms like \( (cx + d) \) at the end to display your derivative in its cleanest, most structured algebraic form.

 

Question 5. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{ax + b}{cx + d} \)
Answer: Let \( f(x) = \frac{ax+b}{cx+d} \).
Applying the quotient rule of differentiation:
\( f'(x) = \frac{(cx+d)\frac{d}{dx}(ax+b) - (ax+b)\frac{d}{dx}(cx+d)}{(cx+d)^2} \)
\( = \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2} \)
\( = \frac{acx + ad - acx - bc}{(cx+d)^2} \)
\( = \frac{ad - bc}{(cx+d)^2} \)
In simple words: The derivative is found using the quotient rule. Calculate the derivative of the top (which is a) and bottom (which is c), plug them into the quotient formula, expand the terms to cancel \( acx \), and get the final result.

Exam Tip: The numerator of the derivative of \( \frac{ax+b}{cx+d} \) is always equal to the determinant value \( ad - bc \), which is a useful shortcut to remember for checking your work.

 

Question 6. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} \)
Answer: Let us define \( f(x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1} \), where \( x \neq 0 \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(x-1)\frac{d}{dx}(x+1) - (x+1)\frac{d}{dx}(x-1)}{(x-1)^2} \), where \( x \neq 0, 1 \)
\( = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} \)
\( = \frac{x - 1 - x - 1}{(x-1)^2} \)
\( = \frac{-2}{(x-1)^2} \), where \( x \neq 0, 1 \)
In simple words: Simplify the fraction first by putting the numerator and denominator over a common denominator x. This turns the equation into \( \frac{x+1}{x-1} \). Then, use the quotient rule to find the rate of change.

Exam Tip: Always state domain restrictions (such as \( x \neq 0, 1 \)) when simplifying fractions to ensure your derivative steps are mathematically sound.

 

Question 7. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{1}{ax^2 + bx + c} \)
Answer: Let \( f(x) = \frac{1}{ax^2 + bx + c} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(ax^2 + bx + c)\frac{d}{dx}(1) - (1)\frac{d}{dx}(ax^2 + bx + c)}{(ax^2 + bx + c)^2} \)
\( = \frac{(ax^2 + bx + c)(0) - 1(2ax + b)}{(ax^2 + bx + c)^2} \)
\( = \frac{-(2ax + b)}{(ax^2 + bx + c)^2} \)
\( = -\frac{2ax + b}{(ax^2 + bx + c)^2} \)
In simple words: Applying the quotient rule, the derivative of the constant on top is 0, and the derivative of the bottom quadratic is \( 2ax+b \). Combining these gives a clean final fraction.

Exam Tip: Alternatively, you can use the chain rule on \( (ax^2+bx+c)^{-1} \) to solve this faster, which gives the exact same result.

 

Question 8. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{ax + b}{px^2 + qx + r} \)
Answer: Let \( f(x) = \frac{ax + b}{px^2 + qx + r} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(px^2 + qx + r)\frac{d}{dx}(ax + b) - (ax + b)\frac{d}{dx}(px^2 + qx + r)}{(px^2 + qx + r)^2} \)
\( = \frac{(px^2 + qx + r)(a) - (ax + b)(2px + q)}{(px^2 + qx + r)^2} \)
\( = \frac{apx^2 + aqx + ar - (2apx^2 + aqx + 2bpx + bq)}{(px^2 + qx + r)^2} \)
\( = \frac{apx^2 + aqx + ar - 2apx^2 - aqx - 2bpx - bq}{(px^2 + qx + r)^2} \)
\( = \frac{-apx^2 - 2bpx + ar - bq}{(px^2 + qx + r)^2} \)
In simple words: Use the quotient rule to expand and differentiate the numerator and denominator. Multiply out the terms in the numerator carefully, group the common terms like \( x^2 \) and \( x \), and write down the simplified expression.

Exam Tip: Take extra care when multiplying binomial expressions in the numerator; distributing the negative sign across all terms inside the parentheses is a frequent place for simple mistakes.

 

Question 9. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{px^2 + qx + r}{ax + b} \)
Answer: Let \( f(x) = \frac{px^2 + qx + r}{ax + b} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(ax + b)\frac{d}{dx}(px^2 + qx + r) - (px^2 + qx + r)\frac{d}{dx}(ax + b)}{(ax + b)^2} \)
\( = \frac{(ax + b)(2px + q) - (px^2 + qx + r)(a)}{(ax + b)^2} \)
\( = \frac{2apx^2 + aqx + 2bpx + bq - (apx^2 + aqx + ar)}{(ax + b)^2} \)
\( = \frac{2apx^2 + aqx + 2bpx + bq - apx^2 - aqx - ar}{(ax + b)^2} \)
\( = \frac{apx^2 + 2bpx + bq - ar}{(ax + b)^2} \)
In simple words: This is the reciprocal shape of the previous question. Use the quotient rule to expand and gather the coefficients of \( x^2 \) and \( x \) in the numerator to simplify your answer.

Exam Tip: After expanding, double-check that similar terms like \( aqx \) cancel out, which helps simplify the numerator significantly.

 

Question 10. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{a}{x^4} - \frac{b}{x^2} + \cos x \)
Answer: Let \( f(x) = \frac{a}{x^4} - \frac{b}{x^2} + \cos x = ax^{-4} - bx^{-2} + \cos x \).
Differentiating with respect to x:
\( f'(x) = \frac{d}{dx}(ax^{-4}) - \frac{d}{dx}(bx^{-2}) + \frac{d}{dx}(\cos x) \)
\( = a(-4x^{-5}) - b(-2x^{-3}) + (-\sin x) \)
\( = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x \)
In simple words: Rewrite fractions as negative powers of x. Then, apply the standard power rule to the first two terms and the trigonometric derivative rule to cosine, which changes into negative sine.

Exam Tip: Converting rational terms with variable denominators into negative exponents makes it much easier to apply the simple power rule directly.

 

Question 11. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( 4\sqrt{x} - 2 \)
Answer: Let \( f(x) = 4\sqrt{x} - 2 \). Differentiating with respect to x:
\( f'(x) = \frac{d}{dx}(4x^{\frac{1}{2}} - 2) \)
\( = \frac{d}{dx}(4x^{\frac{1}{2}}) - \frac{d}{dx}(2) \)
\( = 4\left(\frac{1}{2}x^{-\frac{1}{2}}\right) - 0 \)
\( = 2x^{-\frac{1}{2}} \)
\( = \frac{2}{\sqrt{x}} \)
In simple words: Write the square root of x as x to the power 1/2. When you take the derivative, multiply by 1/2 and subtract 1 from the power, which gives a negative half power, or 2 divided by the square root of x.

Exam Tip: Remember that the derivative of a constant (like 2) is always zero, so you only need to focus on differentiating the variable term.

 

Question 12. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (ax + b)^n \)
Answer: Let us define \( f(x) = (ax + b)^n \). This gives \( f(x+h) = (ax + ah + b)^n \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim_{h \to 0} \frac{(ax + ah + b)^n - (ax+b)^n}{h} \)
\( = (ax+b)^n \lim_{h \to 0} \frac{\left( 1 + \frac{ah}{ax+b} \right)^n - 1}{h} \)
Using the binomial expansion theorem for the fractional term inside the brackets:
\( = (ax+b)^n \lim_{h \to 0} \frac{1}{h} \left[ \left\{ 1 + n\left(\frac{ah}{ax+b}\right) + \frac{n(n-1)}{2!} \left(\frac{ah}{ax+b}\right)^2 + \dots \right\} - 1 \right] \)
\( = (ax+b)^n \lim_{h \to 0} \frac{1}{h} \left[ n\left(\frac{ah}{ax+b}\right) + \frac{n(n-1)}{2} \frac{a^2h^2}{(ax+b)^2} + \dots \right] \)
\( = (ax+b)^n \lim_{h \to 0} \left[ \frac{na}{ax+b} + \frac{n(n-1)}{2} \frac{a^2h}{(ax+b)^2} + \dots \right] \)
Setting the limit as h goes to 0 makes all terms with h vanish:
\( = (ax+b)^n \left[ \frac{na}{ax+b} + 0 \right] \)
\( = na(ax+b)^{n-1} \)
In simple words: This derivative is found using the first principles limit and expanding with the binomial theorem. The terms containing higher powers of h disappear when h goes to 0, which leaves the final rate of change as \( na(ax+b)^{n-1} \).

Exam Tip: If the binomial method is difficult to remember, practice writing out the first three terms of the expansion to clearly show how the h terms cancel out.

 

Question 13. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (ax + b)^n(cx + d)^m \)
Answer: Let \( f(x) = (ax+b)^n (cx+d)^m \).
Using Leibnitz's standard product rule:
\( f'(x) = (ax+b)^n\frac{d}{dx}(cx+d)^m + (cx+d)^m\frac{d}{dx}(ax+b)^n \)

Using the standard derivative results derived in the previous section:
\( \frac{d}{dx}(cx+d)^m = mc(cx+d)^{m-1} \)
\( \frac{d}{dx}(ax+b)^n = na(ax+b)^{n-1} \)

Substituting these values back into the product rule equation:
\( f'(x) = (ax+b)^n \{mc(cx+d)^{m-1}\} + (cx+d)^m \{na(ax+b)^{n-1}\} \)
\( = (ax+b)^{n-1}(cx+d)^{m-1} [mc(ax+b) + na(cx+d)] \)
In simple words: Apply the product rule. Differentiate both exponential terms using the formula proven in the previous question, plug them in, and then pull out the common powers to find the simplified final formula.

Exam Tip: When factoring out common terms in derivatives, always look for the lowest power of each algebraic bracket, which is \( (ax+b)^{n-1} \) and \( (cx+d)^{m-1} \) in this case.

 

Question 14. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \sin(x + a) \)
Answer: Let us define \( f(x) = \sin(x+a) \). This gives \( f(x+h) = \sin(x+h+a) \).
Applying the formula of first principles:
\( f'(x) = \lim_{h \to 0} \frac{\sin(x+h+a) - \sin(x+a)}{h} \)
\( = \lim_{h \to 0} \frac{1}{h} \left[ 2 \cos\left(\frac{x+h+a + x+a}{2}\right) \sin\left(\frac{x+h+a - (x+a)}{2}\right) \right] \) [using \( \sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2} \)]
\( = \lim_{h \to 0} \frac{1}{h} \left[ 2 \cos\left(\frac{2x+2a+h}{2}\right) \sin\left(\frac{h}{2}\right) \right] \)
\( = \lim_{h \to 0} \left[ \cos\left(x + a + \frac{h}{2}\right) \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \)
\( = \lim_{h \to 0} \cos\left(x+a+\frac{h}{2}\right) \cdot \lim_{\frac{h}{2} \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \)
\( = \cos(x+a) \cdot (1) \)
\( = \cos(x+a) \br /> In simple words: This is evaluated using the first principles limit. Use the sine subtraction formula to convert the numerator into a product of cosine and sine, then calculate the standard limit which gives the final derivative \( \cos(x+a) \).

Exam Tip: Since \( a \) is a constant, shifting the sine function horizontally by \( a \) results in a derivative that is similarly shifted cosine function, \( \cos(x+a) \).

 

Question 15. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \csc x \cot x \)
Answer: Let us define \( f(x) = \csc x \cot x \).
Using Leibnitz's standard product rule:
\( f'(x) = \csc x(\cot x)' + \cot x(\csc x)' \) --- (Equation 1)

Using the standard derivative formulas proven in previous first principles sections:
\( (\cot x)' = -\csc^2 x \) --- (Equation 2)
\( (\csc x)' = -\csc x \cot x \) --- (Equation 3)

Substituting Equation 2 and Equation 3 into Equation 1, we obtain:
\( f'(x) = \csc x(-\csc^2 x) + \cot x(-\csc x \cot x) \)
\( = -\csc^3 x - \csc x \cot^2 x \)
\( = -\csc x(\csc^2 x + \cot^2 x) \br /> In simple words: Apply the product rule. Differentiate both cotangent and cosecant terms using their standard derivatives, multiply them by the opposite terms, and factor out the common negative cosecant term.

Exam Tip: You can also rewrite \( \csc x \cot x \) as \( \frac{\cos x}{\sin^2 x} \) and differentiate using the quotient rule to verify your answer.

 

Question 16. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{\cos x}{1 + \sin x} \)
Answer: Let us define \( f(x) = \frac{\cos x}{1 + \sin x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(1 + \sin x)\frac{d}{dx}(\cos x) - (\cos x)\frac{d}{dx}(1 + \sin x)}{(1 + \sin x)^2} \)
\( = \frac{(1 + \sin x)(-\sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2} \)
\( = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} \)
\( = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} \)
Since \( \sin^2 x + \cos^2 x = 1 \), we write:
\( = \frac{-\sin x - 1}{(1 + \sin x)^2} \)
\( = \frac{-(1 + \sin x)}{(1 + \sin x)^2} \)
\( = \frac{-1}{1 + \sin x} \)
In simple words: First apply the quotient rule to differentiate the numerator and denominator. Simplify the resulting equation using the fundamental trigonometric identity \( \sin^2 x + \cos^2 x = 1 \) to cancel out the common bracket from both top and bottom.

Exam Tip: Whenever terms like \( -\sin^2 x - \cos^2 x \) appear, group them together and substitute -1 to easily simplify trigonometric fractions.

 

Question 17. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{\sin x + \cos x}{\sin x - \cos x} \)
Answer: Let us define \( f(x) = \frac{\sin x + \cos x}{\sin x - \cos x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(\sin x - \cos x)\frac{d}{dx}(\sin x + \cos x) - (\sin x + \cos x)\frac{d}{dx}(\sin x - \cos x)}{(\sin x - \cos x)^2} \)
\( = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \)
\( = \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} \)
\( = \frac{-[(\sin^2 x + \cos^2 x - 2\sin x \cos x) + (\sin^2 x + \cos^2 x + 2\sin x \cos x)]}{(\sin x - \cos x)^2} \)
\( = \frac{-[1 - 2\sin x \cos x + 1 + 2\sin x \cos x]}{(\sin x - \cos x)^2} \)
\( = \frac{-2}{(\sin x - \cos x)^2} \)
In simple words: Calculate the derivative using the quotient rule. Expand both binomial squares in the numerator, cancel out the \( 2\sin x\cos x \) terms, and apply the identity \( \sin^2 x + \cos^2 x = 1 \) to find the simplified numerator of -2.

Exam Tip: Watch out for double negative signs in the numerator. Factoring out negative signs early on makes algebraic expansion much easier to track.

 

Question 18. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{\sec x - 1}{\sec x + 1} \)
Answer: Let \( f(x) = \frac{\sec x - 1}{\sec x + 1} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(\sec x + 1)\frac{d}{dx}(\sec x - 1) - (\sec x - 1)\frac{d}{dx}(\sec x + 1)}{(\sec x + 1)^2} \)
\( = \frac{(\sec x + 1)(\sec x \tan x) - (\sec x - 1)(\sec x \tan x)}{(\sec x + 1)^2} \)
Factoring out the term \( \sec x \tan x \) from the numerator:
\( = \frac{\sec x \tan x [(\sec x + 1) - (\sec x - 1)]}{(\sec x + 1)^2} \)
\( = \frac{\sec x \tan x [\sec x + 1 - \sec x + 1]}{(\sec x + 1)^2} \)
\( = \frac{2\sec x \tan x}{(\sec x + 1)^2} \)
In simple words: Differentiate the numerator and denominator directly using the secant derivative rule. Pull out the common factor of \( \sec x\tan x \) from both parts of the numerator, which simplifies the remaining terms to a constant of 2.

Exam Tip: Factoring out common expressions like \( \sec x \tan x \) from the numerator saves you from doing long algebraic expansions and reduces calculation errors.

 

Question 19. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \sin^n x \)
Answer: Let us define \( y = \sin^n x \). We solve this systematically using mathematical induction.

For \( n = 1 \):
\( y = \sin x \implies \frac{dy}{dx} = \cos x \)

For \( n = 2 \):
\( y = \sin^2 x \implies \frac{dy}{dx} = 2\sin x \cos x \)

For \( n = 3 \):
\( y = \sin^3 x \implies \frac{dy}{dx} = 3\sin^2 x \cos x \)

We assert that the general derivative is of the form:
\( \frac{d}{dx}(\sin^n x) = n\sin^{n-1} x \cos x \)

Let our assertion be true for \( n = k \):
\( \frac{d}{dx}(\sin^k x) = k\sin^{k-1} x \cos x \) --- (Equation 1)

Now, consider \( n = k + 1 \):
\( \frac{d}{dx}(\sin^{k+1} x) = \frac{d}{dx}(\sin x \cdot \sin^k x) \)
Applying Leibnitz's product rule of differentiation:
\( = \sin^k x \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(\sin^k x) \)
\( = \sin^k x (\cos x) + \sin x (k\sin^{k-1} x \cos x) \) [using Equation 1]
\( = \cos x \sin^k x + k \sin^k x \cos x \)
\( = (k+1)\sin^k x \cos x \)

Thus, our assertion is true for \( n = k + 1 \).
Hence, by the principle of mathematical induction, the derivative is:
\( \frac{d}{dx}(\sin^n x) = n\sin^{n-1} x \cos x \)
In simple words: This derivative is proven by induction. Since the pattern holds for power 1, 2, and 3, we assume it works for power k, and then use the product rule to prove it successfully holds for power k+1.

Exam Tip: When using mathematical induction for derivative proofs, clearly label the step where you substitute the assumed inductive hypothesis.

 

Question 20. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{a + b\sin x}{c + d\cos x} \)
Answer: Let us define \( f(x) = \frac{a + b\sin x}{c + d\cos x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(c + d\cos x)\frac{d}{dx}(a + b\sin x) - (a + b\sin x)\frac{d}{dx}(c + d\cos x)}{(c + d\cos x)^2} \)
\( = \frac{(c + d\cos x)(b\cos x) - (a + b\sin x)(-d\sin x)}{(c + d\cos x)^2} \)
\( = \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c + d\cos x)^2} \)
\( = \frac{bc\cos x + ad\sin x + bd(\cos^2 x + \sin^2 x)}{(c + d\cos x)^2} \)
Using the standard identity \( \cos^2 x + \sin^2 x = 1 \), we simplify:
\( = \frac{bc\cos x + ad\sin x + bd}{(c + d\cos x)^2} \)
In simple words: Differentiate using the quotient rule. Expand the expressions in the numerator, group the terms containing \( \cos^2 x \) and \( \sin^2 x \) together, and substitute 1 to get the simplified final derivative.

Exam Tip: Always look out for factoring opportunities like \( bd(\cos^2 x + \sin^2 x) \) to shrink trigonometric expressions into constants.

 

Question 21. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{\sin(x+a)}{\cos x} \)
Answer: Let \( f(x) = \frac{\sin(x+a)}{\cos x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{\cos x \frac{d}{dx}\sin(x+a) - \sin(x+a)\frac{d}{dx}\cos x}{\cos^2 x} \)
\( = \frac{\cos x \cos(x+a) - \sin(x+a)(-\sin x)}{\cos^2 x} \)
\( = \frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x} \)
Using the standard identity \( \cos(A-B) = \cos A\cos B + \sin A\sin B \):
\( = \frac{\cos(x+a-x)}{\cos^2 x} \)
\( = \frac{\cos a}{\cos^2 x} \)
In simple words: First apply the quotient rule to differentiate both trigonometric terms. This gives an expression in the numerator that matches the formula for \( \cos(A-B) \). Simplifying this cancels out the variable x from the top, leaving \( \cos a \) divided by \( \cos^2 x \).

Exam Tip: Recognize compound angle identities like \( \cos A\cos B + \sin A\sin B = \cos(A-B) \) to simplify complicated trigonometric fractions into constants.

 

Question 22. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( x^4(5\sin x - 3\cos x) \)
Answer: Let us define \( f(x) = x^4(5\sin x - 3\cos x) \).
Using the product rule of derivatives:
\( f'(x) = x^4 \frac{d}{dx}(5\sin x - 3\cos x) + (5\sin x - 3\cos x) \frac{d}{dx}(x^4) \)
\( = x^4(5\cos x - 3(-\sin x)) + (5\sin x - 3\cos x)(4x^3) \)
\( = x^4(5\cos x + 3\sin x) + 4x^3(5\sin x - 3\cos x) \)
Factoring out the term \( x^3 \):
\( = x^3[x(5\cos x + 3\sin x) + 4(5\sin x - 3\cos x)] \)
\( = x^3[5x\cos x + 3x\sin x + 20\sin x - 12\cos x] \)
In simple words: Differentiate using the product rule. Work out the derivatives of both components, multiply across the opposite terms, and factor out \( x^3 \) at the end to get the simplified answer.

Exam Tip: Pulling out common power factors like \( x^3 \) early in the simplification step helps keep the remaining algebraic terms manageable and neat.

 

Question 23. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (x^2 + 1)\cos x \)
Answer: Let \( f(x) = (x^2 + 1)\cos x \).
Using the product rule of derivatives:
\( f'(x) = (x^2 + 1)\frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(x^2+1) \)
\( = (x^2 + 1)(-\sin x) + \cos x (2x) \)
\( = -(x^2 + 1)\sin x + 2x\cos x \)
In simple words: Differentiate using the product rule. Differentiate \( x^2+1 \) to get \( 2x \), and differentiate \( \cos x \) to get \( -\sin x \). Multiply them across and sum them up to get the final derivative.

Exam Tip: Keep a close eye on the negative sign when differentiating cosine to avoid sign errors in the final sum.

 

Question 24. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (ax^2 + \sin x)(p + q\cos x) \)
Answer: Let \( f(x) = (ax^2 + \sin x)(p + q\cos x) \).
Using the product rule of derivatives:
\( f'(x) = (ax^2 + \sin x)\frac{d}{dx}(p + q\cos x) + (p + q\cos x)\frac{d}{dx}(ax^2 + \sin x) \)
\( = (ax^2 + \sin x)(0 - q\sin x) + (p + q\cos x)(2ax + \cos x) \)
\( = -q\sin x(ax^2 + \sin x) + (p + q\cos x)(2ax + \cos x) \)
In simple words: Apply the product rule. Calculate the derivative of both brackets, cross-multiply them by the original expressions, and sum them together to get the final derivative.

Exam Tip: Since both brackets contain a mix of polynomial and trigonometric functions, keep them factored in parenthesis rather than doing long term expansions.

 

Question 25. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (x + \cos x)(x - \tan x) \)
Answer: Let \( f(x) = (x + \cos x)(x - \tan x) \).
Using the product rule of derivatives:
\( f'(x) = (x + \cos x)\frac{d}{dx}(x - \tan x) + (x - \tan x)\frac{d}{dx}(x + \cos x) \)
\( = (x + \cos x)(1 - \sec^2 x) + (x - \tan x)(1 - \sin x) \)
Using the standard identity \( 1 - \sec^2 x = -\tan^2 x \), we simplify:
\( = -\tan^2 x(x + \cos x) + (x - \tan x)(1 - \sin x) \)
In simple words: Apply the product rule. Work out the derivatives of both parts, using the fact that the derivative of tangent is secant squared, and substitute the identity \( 1 - \sec^2 x = -\tan^2 x \) to get the final answer.

Exam Tip: Remember the basic Pythagorean identity \( 1 + \tan^2 x = \sec^2 x \) which is frequently used to simplify terms in derivatives of tangent functions.

 

Question 26. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{4x + 5\sin x}{3x + 7\cos x} \)
Answer: Let \( f(x) = \frac{4x + 5\sin x}{3x + 7\cos x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(3x + 7\cos x)\frac{d}{dx}(4x + 5\sin x) - (4x + 5\sin x)\frac{d}{dx}(3x + 7\cos x)}{(3x + 7\cos x)^2} \)
\( = \frac{(3x + 7\cos x)(4 + 5\cos x) - (4x + 5\sin x)(3 - 7\sin x)}{(3x + 7\cos x)^2} \)
\( = \frac{12x + 15x\cos x + 28\cos x + 35\cos^2 x - [12x - 28x\sin x + 15\sin x - 35\sin^2 x]}{(3x + 7\cos x)^2} \)
\( = \frac{12x + 15x\cos x + 28\cos x + 35\cos^2 x - 12x + 28x\sin x - 15\sin x + 35\sin^2 x}{(3x + 7\cos x)^2} \)
\( = \frac{15x\cos x + 28x\sin x + 28\cos x - 15\sin x + 35(\cos^2 x + \sin^2 x)}{(3x + 7\cos x)^2} \)
Since \( \cos^2 x + \sin^2 x = 1 \):
\( = \frac{35 + 15x\cos x + 28x\sin x + 28\cos x - 15\sin x}{(3x + 7\cos x)^2} \)
In simple words: Apply the quotient rule to expand and differentiate both parts of the fraction. Multiply out the long brackets in the numerator, cancel out the matching 12x terms, group the sine squared and cosine squared terms, and simplify using standard trigonometric identities.

Exam Tip: Keep your bracket expansions tidy and organized to avoid sign errors when distributing the negative sign over the second half of the quotient rule numerator.

 

Question 27. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{x^2\cos\left(\frac{\pi}{4}\right)}{\sin x} \)
Answer: Let \( f(x) = \frac{x^2\cos\left(\frac{\pi}{4}\right)}{\sin x} \).
Since \( \cos\left(\frac{\pi}{4}\right) \) is a constant real number multiplier, we can pull it outside of the differentiation:
Applying the quotient rule of derivatives:
\( f'(x) = \cos\left(\frac{\pi}{4}\right) \left[ \frac{\sin x \frac{d}{dx}(x^2) - x^2\frac{d}{dx}(\sin x)}{\sin^2 x} \right] \)
\( = \cos\left(\frac{\pi}{4}\right) \left[ \frac{\sin x (2x) - x^2\cos x}{\sin^2 x} \right] \)
\( = \cos\left(\frac{\pi}{4}\right) \frac{x[2\sin x - x\cos x]}{\sin^2 x} \)
In simple words: Treat the \( \cos(\pi/4) \) term as a constant multiplier. Differentiate the remaining fraction \( x^2/\sin x \) using the quotient rule, factor out the common x from the numerator, and re-multiply by the constant.

Exam Tip: Never waste time trying to apply derivative rules to constant terms like \( \cos\left(\frac{\pi}{4}\right) \); treating them as simple scalar multipliers simplifies your calculations.

 

Question 28. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{x}{1 + \tan x} \)
Answer: Let \( f(x) = \frac{x}{1 + \tan x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{(1 + \tan x)\frac{d}{dx}(x) - x\frac{d}{dx}(1 + \tan x)}{(1 + \tan x)^2} \)
\( = \frac{(1 + \tan x)(1) - x(\sec^2 x)}{(1 + \tan x)^2} \)
\( = \frac{1 + \tan x - x\sec^2 x}{(1 + \tan x)^2} \)
In simple words: Differentiate using the quotient rule. The derivative of the top term x is 1, and the derivative of the bottom term \( 1+\tan x \) is \( \sec^2 x \). Combine them into the quotient formula to find the derivative.

Exam Tip: Remember that the derivative of a sum \( 1 + \tan x \) is simply the derivative of each individual term, which yields \( 0 + \sec^2 x \).

 

Question 29. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( (x + \sec x)(x - \tan x) \)
Answer: Let \( f(x) = (x + \sec x)(x - \tan x) \).
Using the product rule of derivatives:
\( f'(x) = (x + \sec x)\frac{d}{dx}(x - \tan x) + (x - \tan x)\frac{d}{dx}(x + \sec x) \)
\( = (x + \sec x)(1 - \sec^2 x) + (x - \tan x)(1 + \sec x \tan x) \)
Since \( 1 - \sec^2 x = -\tan^2 x \), we can also write:
\( = -\tan^2 x(x + \sec x) + (x - \tan x)(1 + \sec x \tan x) \)
In simple words: Apply the product rule to differentiate both expressions. Remember that the derivative of \( \sec x \) is \( \sec x \tan x \) and the derivative of \( \tan x \) is \( \sec^2 x \). Use trigonometric identities to write down the final expression.

Exam Tip: Leaving the derivative in its factored form is completely acceptable and preferred by examiners over long, complicated expansions.

 

Question 30. Find the derivative of the following functions (it is to be understood that \( a, b, c, d, p, q, r \) and \( s \) are fixed non-zero constants and \( m \) and \( n \) are integers): \( \frac{x}{\sin^n x} \)
Answer: Let \( f(x) = \frac{x}{\sin^n x} \).
Applying the quotient rule of derivatives:
\( f'(x) = \frac{\sin^n x \frac{d}{dx}(x) - x\frac{d}{dx}(\sin^n x)}{\sin^{2n} x} \)

Using the derivative result for \( \sin^n x \) derived in Question 19:
\( \frac{d}{dx}(\sin^n x) = n\sin^{n-1} x \cos x \)

Substituting this back into our quotient rule equation:
\( f'(x) = \frac{\sin^n x(1) - x(n\sin^{n-1} x \cos x)}{\sin^{2n} x} \)
\( = \frac{\sin^{n-1} x [\sin x - nx\cos x]}{\sin^{2n} x} \)
\( = \frac{\sin x - nx\cos x}{\sin^{n+1} x} \)
In simple words: Use the quotient rule to differentiate both parts. Use the power rule on \( \sin^n x \) to get \( n\sin^{n-1} x \cos x \). Then, pull out the common factor \( \sin^{n-1} x \) from the top to cancel it with the bottom, leaving a simplified fraction.

Exam Tip: Always look to cancel common powers of \( \sin x \) between the numerator and denominator to simplify rational trigonometric derivatives to their simplest forms.