Question

If $y=f\left(x\right)$ is a differentiable function of $x$, then show that
$\frac{d^{2}x}{d{y}^2}=-{\left(\frac{dy}{dx}\right)}^{-3}.\frac{d^{2}y}{d{x}^2}$

Answer 1

If $y=f\left(x\right)$ is a differentiable function of x such that inverse function $x=f^{-1}\left(y\right)$ exists, then $\frac{dx}{dy}=\frac{1}{\left(\frac{dy}{dx}\right)}$. where $\frac{dy}{dx}\ne 0$
$\therefore \frac{d^{2}x}{d{y}^2}=\frac{d}{dx}\left(\frac{dx}{dy}\right)$
$=\frac{d}{dy}\left[\frac{1}{\left(\frac{dy}{dx}\right)}\right]$
$=\frac{d}{Dx}{\left(\frac{dy}{dx}\right)}^{-1}\times \frac{dx}{dy}$
$=-1{\left(\frac{dy}{dx}\right)}^{-2}\cdot \frac{d}{dx}\left(\frac{dy}{dx}\right)\times \frac{1}{\left(\frac{dy}{dx}\right)}$
$=-{\left(\frac{dy}{dx}\right)}^{-2}\cdot \frac{d^{2}y}{d{x}^2}\cdot {\left(\frac{dy}{dx}\right)}^{-1}$
$\therefore \frac{d^{2}x}{d{y}^2}=-{\left(\frac{dy}{dx}\right)}^{-3}\cdot \frac{d^{2}y}{d{x}^2}$