Question

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

• A. $\frac{64}{3}$
• B. $\frac{16}{3}$
• C. $\frac{3}{16}$
• D. $\frac{3}{64}$

Answer 1

The Correct Option is D

$\because \frac{d^{2} x}{d y^{2}} =\frac{d}{d y}\left(\frac{d x}{d y}\right)$
$=\frac{d x}{d y} \frac{d}{d x}\left(\frac{1}{\left(\frac{d y}{d x}\right)}\right)$
$=\left(\frac{d x}{d y}\right)\left(\frac{-\frac{d^{2} y}{d x^{2}}}{\left(\frac{d y}{d x}\right)^{2}}\right)$
$=-\frac{\left(\frac{d^{2} y}{d x^{2}}\right)}{\left(\frac{d y}{d x}\right)^{3}}$
$\therefore\left(\frac{d^{2} x}{d y^{2}}\right)_{p}=-\frac{(-3)}{(4)^{3}}=\frac{3}{64}$