Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let y=f(x)y = f(x) be a real valued twice differentiable function defined on R\mathbb{R}. Then d2ydx2(dxdy)3+d2xdy2\dfrac{d^2 y}{dx^2}\left(\dfrac{dx}{dy}\right)^3 + \dfrac{d^2 x}{dy^2} is
Adydx\dfrac{dy}{dx}
Bdxdy\dfrac{dx}{dy}
Cd2ydx2\dfrac{d^2 y}{dx^2}
D00correct
Solution
Step 1: Express the second derivative through the inverse relation Since dydx=1dxdy\frac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}:
d2ydx2=ddx(dxdy)1=(dxdy)2ddx(dxdy)\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{dx}{dy}\right)^{-1} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dx}\left(\frac{dx}{dy}\right)
Step 2: Convert the inner derivative to yy
ddx(dxdy)=d2xdy2dydx=d2xdy21dxdy\frac{d}{dx}\left(\frac{dx}{dy}\right) = \frac{d^2 x}{dy^2}\cdot\frac{dy}{dx} = \frac{d^2 x}{dy^2}\cdot\frac{1}{\frac{dx}{dy}}
Step 3: Combine
d2ydx2=d2xdy2(dxdy)3    d2ydx2(dxdy)3+d2xdy2=0\frac{d^2 y}{dx^2} = -\frac{\frac{d^2 x}{dy^2}}{\left(\frac{dx}{dy}\right)^3} \implies \frac{d^2 y}{dx^2}\left(\frac{dx}{dy}\right)^3 + \frac{d^2 x}{dy^2} = 0
Answer: (4)
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