2 marks, −0.667MathematicsLimits, continuity and differentiabilityImplicit, logarithmic and higher-order differentiationModerateabout 2 min

IIT JAM 2022 Question 11: Limits, continuity and differentiability

Let $x^3 + 3y^2 = 4$ for all $x, y \in \mathbb{R}$ , $y' = \dfrac{dy}{dx}$ and $y'' = \dfrac{d^2y}{dx^2}$ . Then

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Answer and worked solution

Answer: (D)

Differentiate $x^3 + 3y^2 = 4$ once: $3x^2 + 6y\,y' = 0$ , i.e. $x^2 + 2y\,y' = 0$ .

Differentiate again: $2x + 2\left[(y')^2 + y\,y''\right] = 0$ , i.e. $x + (y')^2 + y\,y'' = 0$ . This is option (D).

The trap: (A)

x^2 + y\,y'' + (y')^2 = 0

Option (A) keeps an $x^2$ term from the first differentiation instead of the $x$ that appears after differentiating a second time; only the second derivative relation is asked.

Plain-text version

Let x³ + 3y² = 4 for all x, y ∈ ℝ, y′ = dy/dx and y′′ = d²y/dx². Then (A) x² + y y′′ + (y′)² = 0 (B) 2x + y′′ + 2(y′)² = 0 (C) x + (y′)² = 0 (D) x + y y′′ + (y′)² = 0

Also written as: Let x^3 + 3y^2 = 4 for all x, y in R, y' = dy/dx and y'' = d^2y/dx^2. Then

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What this question tests

This question is on Implicit, logarithmic and higher-order differentiation, part of Limits, continuity and differentiability in Mathematics.

Differentiating relations that are not solved for y, and taking second and higher derivatives.

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