IIT-JEE 2007 · Paper 2 · Mathematics Q48

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MathematicsSingle correct+3 / -1Question 48 of 66

$\dfrac{d^2x}{dy^2}$ equals

\left(\dfrac{d^2y}{dx^2}\right)^{-1}

-\left(\dfrac{d^2y}{dx^2}\right)^{-1}\left(\dfrac{dy}{dx}\right)^{-3}

\left(\dfrac{d^2y}{dx^2}\right)\left(\dfrac{dy}{dx}\right)^{-2}

-\left(\dfrac{d^2y}{dx^2}\right)\left(\dfrac{dy}{dx}\right)^{-3}

Answer and worked solution

Official answer: D

First derivative. $\dfrac{dx}{dy}=\left(\dfrac{dy}{dx}\right)^{-1}$ .

Second derivative (chain rule):
$\frac{d^2x}{dy^2}=\frac d{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right]\frac{dx}{dy}=-\left(\frac{dy}{dx}\right)^{-2}\frac{d^2y}{dx^2}\cdot\left(\frac{dy}{dx}\right)^{-1}=-\frac{d^2y}{dx^2}\left(\frac{dy}{dx}\right)^{-3}.$

(D)

Topics:Higher-order derivatives Chain rule and implicit differentiation

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IIT-JEE 2007 · Paper 2, Mathematics question 48