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

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Aicharya · Accepted Answer

$\begin{array}{ll} \because \frac{d^2 x}{d y^2} =\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d x}{d y} \frac{d}{d x}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right) \ =\left(\frac{d x}{d y}\right)\left(\frac{-\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^2}\right)=-\frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d y}{d x}\right)^3} \ \therefore\left(\frac{d^2 x}{d y^2}\right)_p =-\frac{(-3)}{(4)^3}=\frac{3}{64} \end{array}$ Hence, option (4) is correct.

If \(y=f(x)\) is twice differentiable function such that at a point \(P, \frac{d y}{d x}=4, \frac{d^2 y}{d…

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