Method of DifferentiationmediumFree

Method of Differentiation: Let Real Valued Twice Differentiable Function Defined

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

Question

Let

y = f(x)

be a real valued twice differentiable function defined on

\mathbb{R}

. Then

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

\dfrac{dy}{dx}

\dfrac{dx}{dy}

\dfrac{d^2 y}{dx^2}

0

correct

Solution

Step 1: Express the second derivative through the inverse relation Since

\frac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}

\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

y

\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

\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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