If y=f(x) is a twice differentiable function | vedclass

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(D) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.Now,differentiating both sides with respect to $y$ using the chain rule:$\frac{d^2x}{dy^2}

Vedclass · Accepted Answer

$\frac{3}{64}$

Vedclass · Answer

(D) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.Now,differentiating both sides with respect to $y$ using the chain rule:$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{\frac{dy}{dx}} \right) = \frac{d}{dx} \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{dx}{dy}$.Using the quotient rule or power rule:$\frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right) = -\left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2}$.Substituting this back:$\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2} \cdot \frac{1}{\frac{dy}{dx}} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}$.Given $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at point $P$:$\left( \frac{d^2x}{dy^2} \right)_P = -\frac{-3}{(4)^3} = \frac{3}{64}$.Thus,option $D$ is correct.

If $y=f(x)$ is a twice differentiable function such that at a point $P$,$\frac{dy}{dx}=4$ and $\frac{d^2y}{dx^2}=-3$,then $\left(\frac{d^2x}{dy^2}\right)_P$ is equal to:

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