frac{d^2x}{dy^2} = | vedclass

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Question

(C) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.Now,differentiate with respect to $y$:$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.Now,differentiate with respect to $y$:$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dy} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.Using the chain rule,$\frac{d}{dy} = \frac{dx}{dy} \cdot \frac{d}{dx} = \frac{1}{\frac{dy}{dx}} \cdot \frac{d}{dx}$:$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.Applying the power rule:$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \left( -1 \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} \right)$.Simplifying the expression:$\frac{d^2x}{dy^2} = - \left( \frac{dy}{dx} \right)^{-1} \cdot \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = - \left( \frac{dy}{dx} \right)^{-3} \left( \frac{d^2y}{dx^2} \right)$.

$\frac{d^2x}{dy^2} = $

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