Maharashtra Board Class 12 Maths Part 1 Chapter 3 Differentiation 3.4 Solutions

1. Find \( \frac{dy}{dx} \) if:

Question 1. \( \sqrt{x} + \sqrt{y} = \sqrt{a} \)
Answer:
Solution:
\( \sqrt{x} + \sqrt{y} = \sqrt{a} \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
\( \implies \frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = -\frac{2\sqrt{y}}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = -\sqrt{\frac{y}{x}} \)
In simple words: To find dy/dx, differentiate the equation term by term with respect to x, remembering to apply the chain rule for terms involving y, then isolate dy/dx.

🎯 Exam Tip: Remember the derivative of \( \sqrt{f(x)} \) is \( \frac{1}{2\sqrt{f(x)}} f'(x) \). This is a crucial step in such problems.

Question 2. \( x^3 + y^3 + 4x^3y = 0 \)
Answer:
Solution:
\( x^3 + y^3 + 4x^3y = 0 \)
Differentiating both sides w.r.t. x, we get
\( 3x^2 + 3y^2 \frac{dy}{dx} + 4 \left[ x^3 \frac{dy}{dx} + y \frac{d}{dx}(x^3) \right] = 0 \)
\( \implies 3x^2 + 3y^2 \frac{dy}{dx} + 4x^3 \frac{dy}{dx} + 4y \cdot 3x^2 = 0 \)
\( \implies 3y^2 \frac{dy}{dx} + 4x^3 \frac{dy}{dx} = -3x^2 - 12x^2y \)
\( \implies (3y^2 + 4x^3) \frac{dy}{dx} = -3x^2(1+4y) \)
\( \implies \frac{dy}{dx} = -\frac{3x^2(1+4y)}{3y^2 + 4x^3} \)
In simple words: Differentiate each term of the given implicit equation with respect to x, applying the product rule for \( 4x^3y \) and the chain rule for terms involving y, then algebraically solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Pay close attention to the product rule application for terms like \( 4x^3y \), as it's a common area for errors in implicit differentiation.

Question 3. \( x^3 + x^2y + xy^2 + y^3 = 81 \)
Answer:
Solution:
\( x^3 + x^2y + xy^2 + y^3 = 81 \)
Differentiating both sides w.r.t. x, we get
\( 3x^2 + \left[ x^2 \frac{dy}{dx} + y \frac{d}{dx}(x^2) \right] + \left[ x \frac{d}{dx}(y^2) + y^2 \frac{d}{dx}(x) \right] + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies 3x^2 + x^2 \frac{dy}{dx} + y \cdot 2x + x \cdot 2y \frac{dy}{dx} + y^2 \cdot 1 + 3y^2 \frac{dy}{dx} = 0 \)
\( \implies x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -3x^2 - 2xy - y^2 \)
\( \implies (x^2 + 2xy + 3y^2) \frac{dy}{dx} = -(3x^2 + 2xy + y^2) \)
\( \implies \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2} \)
In simple words: Apply differentiation rules, including the product rule for \( x^2y \) and \( xy^2 \), and the chain rule for \( y^3 \), to each term, then collect all terms with \( \frac{dy}{dx} \) to one side and solve.

🎯 Exam Tip: Grouping terms with \( \frac{dy}{dx} \) accurately after differentiation is critical for simplifying the expression and arriving at the correct answer.

2. Find \( \frac{dy}{dx} \) if:

Question 1. \( y \cdot e^x + x \cdot e^y = 1 \)
Answer:
Solution:
\( y \cdot e^x + x \cdot e^y = 1 \)
Differentiating both sides w.r.t. x, we get
\( \frac{d}{dx}(y e^x) + \frac{d}{dx}(x e^y) = 0 \)
\( \implies y \frac{d}{dx}(e^x) + e^x \frac{d}{dx}(y) + x \frac{d}{dx}(e^y) + e^y \frac{d}{dx}(x) = 0 \)
\( \implies y e^x + e^x \frac{dy}{dx} + x e^y \frac{dy}{dx} + e^y \cdot 1 = 0 \)
\( \implies (e^x + x e^y) \frac{dy}{dx} = -e^y - y e^x \)
\( \implies \frac{dy}{dx} = -\frac{e^y + y e^x}{e^x + x e^y} \)
In simple words: Use the product rule for both \( y \cdot e^x \) and \( x \cdot e^y \), applying the chain rule for \( e^y \), then rearrange the equation to isolate \( \frac{dy}{dx} \).

🎯 Exam Tip: Remember that \( \frac{d}{dx}(e^y) = e^y \frac{dy}{dx} \). Incorrectly differentiating exponential terms with respect to y is a common mistake.

Question 2. \( x^y = e^{(x-y)} \)
Answer:
Solution:
\( x^y = e^{(x-y)} \)
Taking logarithm on both sides, we get
\( \log (x^y) = \log (e^{(x-y)}) \)
\( \implies y \log x = (x-y) \log e \)
\( \implies y \log x = x-y \quad [: \log e = 1] \)
\( \implies y + y \log x = x \)
\( \implies y(1 + \log x) = x \)
\( \implies y = \frac{x}{1+\log x} \)
Now, differentiating with respect to x, we get
\( \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{1+\log x} \right) \)
\( \implies \frac{dy}{dx} = \frac{(1+\log x) \frac{d}{dx}(x) - x \frac{d}{dx}(1+\log x)}{(1+\log x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1+\log x) \cdot 1 - x \left( 0+\frac{1}{x} \right)}{(1+\log x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1+\log x - 1}{(1+\log x)^2} \)
\( \implies \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2} \)
In simple words: Take the natural logarithm of both sides to simplify the exponent, rearrange to solve for y, then differentiate the resulting explicit function of y using the quotient rule.

🎯 Exam Tip: Logarithmic differentiation is key for expressions with variables in the exponent. Simplification using log properties before differentiating reduces complexity significantly.

Question 3. \( xy = \log(xy) \)
Answer:
Solution:
\( xy = \log(xy) \)
Using logarithm properties, we can write
\( xy = \log x + \log y \)
Differentiating both sides w.r.t. x, we get
\( x \frac{dy}{dx} + y \frac{d}{dx}(x) = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \)
\( \implies x \frac{dy}{dx} + y \cdot 1 = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \)
\( \implies x \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - y \)
\( \implies \left( x - \frac{1}{y} \right) \frac{dy}{dx} = \frac{1-xy}{x} \)
\( \implies \left( \frac{xy-1}{y} \right) \frac{dy}{dx} = \frac{1-xy}{x} \)
\( \implies \frac{-(1-xy)}{y} \frac{dy}{dx} = \frac{1-xy}{x} \)
\( \implies -\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \)
\( \implies \frac{dy}{dx} = -\frac{y}{x} \)
In simple words: First, use logarithm properties to expand \( \log(xy) \) into \( \log x + \log y \). Then, implicitly differentiate both sides with respect to x, applying the product rule and chain rule as needed, and finally solve for \( \frac{dy}{dx} \).

🎯 Exam Tip: Recognizing that \( \log(xy) = \log x + \log y \) before differentiating simplifies the problem significantly and prevents more complex chain rule applications.

3. Solve The Following:

Question 1. If \( x^5 \cdot y^7 = (x + y)^{12} \), then show that \( \frac{dy}{dx} = \frac{y}{x} \)
Answer:
Solution:
Given \( x^5 \cdot y^7 = (x + y)^{12} \)
Taking logarithm on both sides,
\( \log(x^5 \cdot y^7) = \log((x + y)^{12}) \)
\( \implies \log x^5 + \log y^7 = 12 \log(x + y) \)
\( \implies 5 \log x + 7 \log y = 12 \log(x + y) \)
Differentiating both sides w.r.t. x, we get
\( 5 \frac{1}{x} + 7 \frac{1}{y} \frac{dy}{dx} = 12 \frac{1}{x+y} \frac{d}{dx}(x+y) \)
\( \implies \frac{5}{x} + \frac{7}{y} \frac{dy}{dx} = \frac{12}{x+y} \left( 1 + \frac{dy}{dx} \right) \)
\( \implies \frac{5}{x} + \frac{7}{y} \frac{dy}{dx} = \frac{12}{x+y} + \frac{12}{x+y} \frac{dy}{dx} \)
\( \implies \frac{7}{y} \frac{dy}{dx} - \frac{12}{x+y} \frac{dy}{dx} = \frac{12}{x+y} - \frac{5}{x} \)
\( \implies \left( \frac{7}{y} - \frac{12}{x+y} \right) \frac{dy}{dx} = \frac{12x - 5(x+y)}{x(x+y)} \)
\( \implies \left( \frac{7(x+y) - 12y}{y(x+y)} \right) \frac{dy}{dx} = \frac{12x - 5x - 5y}{x(x+y)} \)
\( \implies \left( \frac{7x+7y - 12y}{y(x+y)} \right) \frac{dy}{dx} = \frac{7x - 5y}{x(x+y)} \)
\( \implies \left( \frac{7x - 5y}{y(x+y)} \right) \frac{dy}{dx} = \frac{7x - 5y}{x(x+y)} \)
\( \implies \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{y}{x} \)
In simple words: Take the logarithm of both sides to simplify the exponents, apply log properties to expand the terms, then differentiate implicitly with respect to x, and solve algebraically to prove \( \frac{dy}{dx} = \frac{y}{x} \).

🎯 Exam Tip: Logarithmic differentiation is essential here. Be careful with algebraic simplification after differentiation, ensuring all terms are handled correctly to reach the desired result.

Question 2. If \( \log(x + y) = \log(xy) + a \), then show that \( \frac{dy}{dx} = -\frac{y^2}{x^2} \)
Answer:
Solution:
Given \( \log(x + y) = \log(xy) + a \)
Using logarithm properties, we can write
\( \log(x + y) = \log x + \log y + a \)
Differentiating both sides w.r.t. x, we get
\( \frac{1}{x+y} \frac{d}{dx}(x+y) = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} + 0 \)
\( \implies \frac{1}{x+y} \left( 1 + \frac{dy}{dx} \right) = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \)
\( \implies \frac{1}{x+y} + \frac{1}{x+y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \)
\( \implies \frac{1}{x+y} \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - \frac{1}{x+y} \)
\( \implies \left( \frac{1}{x+y} - \frac{1}{y} \right) \frac{dy}{dx} = \frac{1}{x} - \frac{1}{x+y} \)
\( \implies \left( \frac{y - (x+y)}{y(x+y)} \right) \frac{dy}{dx} = \frac{(x+y) - x}{x(x+y)} \)
\( \implies \left( \frac{y - x - y}{y(x+y)} \right) \frac{dy}{dx} = \frac{x+y - x}{x(x+y)} \)
\( \implies \left( \frac{-x}{y(x+y)} \right) \frac{dy}{dx} = \frac{y}{x(x+y)} \)
\( \implies \frac{-x}{y} \frac{dy}{dx} = \frac{y}{x} \)
\( \implies \frac{dy}{dx} = -\frac{y^2}{x^2} \)
In simple words: Simplify the given logarithmic equation using log properties, then implicitly differentiate each term with respect to x. Combine terms containing \( \frac{dy}{dx} \) and solve for it to arrive at \( -\frac{y^2}{x^2} \).

🎯 Exam Tip: Pay careful attention to the common denominator when combining fractions after differentiation, as algebraic errors here can lead to an incorrect final result.

Question 3. If \( e^x + e^y = e^{(x+y)} \), then show that \( \frac{dy}{dx} = -e^{y-x} \).
Answer:
Solution:
Given \( e^x + e^y = e^{(x+y)} \) ..........(1)
Differentiating both sides w.r.t. x, we get
\( e^x + e^y \frac{dy}{dx} = e^{(x+y)} \frac{d}{dx}(x+y) \)
\( \implies e^x + e^y \frac{dy}{dx} = e^{(x+y)} \left( 1 + \frac{dy}{dx} \right) \)
\( \implies e^x + e^y \frac{dy}{dx} = e^{(x+y)} + e^{(x+y)} \frac{dy}{dx} \)
\( \implies e^y \frac{dy}{dx} - e^{(x+y)} \frac{dy}{dx} = e^{(x+y)} - e^x \)
\( \implies [e^y - e^{(x+y)}] \frac{dy}{dx} = e^{(x+y)} - e^x \)
From (1), we know \( e^{(x+y)} = e^x + e^y \). Substitute this back:
\( \implies [e^y - (e^x + e^y)] \frac{dy}{dx} = (e^x + e^y) - e^x \)
\( \implies [e^y - e^x - e^y] \frac{dy}{dx} = e^y \)
\( \implies -e^x \frac{dy}{dx} = e^y \)
\( \implies \frac{dy}{dx} = -\frac{e^y}{e^x} \)
\( \implies \frac{dy}{dx} = -e^{y-x} \)
In simple words: Differentiate the given equation implicitly with respect to x. Substitute the original equation back into the differentiated form to simplify the expression, and then solve for \( \frac{dy}{dx} \) to get \( -e^{y-x} \).

🎯 Exam Tip: The crucial step here is substituting the original equation \( e^x + e^y = e^{(x+y)} \) back into the differentiated expression. This substitution simplifies the algebra significantly towards the final proof.