JEE Challenger
More from Continuity and Differentiability
MathMedAdv 2007 • Paper 2

Second Derivative of Inverse Function

d2xdy2\frac{d^2x}{dy^2} equals

Options

A

(d2ydx2)−1\left(\frac{d^2y}{dx^2}\right)^{-1}

B

−(d2ydx2)−1(dydx)−3-\left(\frac{d^2y}{dx^2}\right)^{-1}\left(\frac{dy}{dx}\right)^{-3}

C

(d2ydx2)(dydx)−2\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-2}

D

−(d2ydx2)(dydx)−3-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}

Correct

Topics & Concepts

Continuity and DifferentiabilityHigher Order DerivativesInverse Function DerivativeChain Rule

Step-by-Step Solution

We know that the first derivative of the inverse function is:
dxdy=(dydx)−1\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}

Differentiating with respect to yy using the chain rule:
d2xdy2=ddy[(dydx)−1]=dxdy⋅ddx[(dydx)−1]\frac{d^2x}{dy^2} = \frac{d}{dy}\left[\left(\frac{dy}{dx}\right)^{-1}\right] = \frac{dx}{dy} \cdot \frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right]

Using the power rule on (dydx)−1\left(\frac{dy}{dx}\right)^{-1}:
ddx[(dydx)−1]=−(dydx)−2d2ydx2\frac{d}{dx}\left[\left(\frac{dy}{dx}\right)^{-1}\right] = -\left(\frac{dy}{dx}\right)^{-2} \frac{d^2y}{dx^2}

Substituting dxdy=(dydx)−1\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}:
d2xdy2=(dydx)−1[−(dydx)−2d2ydx2]=−(d2ydx2)(dydx)−3\frac{d^2x}{dy^2} = \left(\frac{dy}{dx}\right)^{-1} \left[ -\left(\frac{dy}{dx}\right)^{-2} \frac{d^2y}{dx^2} \right] = -\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}

Correct Answer: D

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