Question

If $y=f\left(x\right)$ is a differentiable function of $x$ such that inverse function $x=f^{-1}\left(y\right)$ exists, then prove that $x$ is a differentiable function of $y$ and $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$ where $\frac{dy}{dx}\ne 0$.
Hence find $\frac{d}{dx}\left({\tan }^{-1}x\right)$

Answer 1

Given $x=f^{-1}\left(y\right)$

Differentiate both sides w.r.t. $x$, we get

$\frac{dx}{dx}=\frac{d\left(f^{-1}\right(y\left)\right)}{dx}$

$\Rightarrow 1=\frac{d\left(f^{-1}\right(y\left)\right)}{dx}$

Apply chain rule to the RHS,

$1=\frac{d\left(f^{-1}\right(y\left)\right)}{dy}\cdot \frac{dy}{dx}$
$\Rightarrow \frac{1}{\frac{dy}{dx}}=\frac{d\left(f^{-1}\right(y\left)\right)}{dy}$

$\Rightarrow \frac{1}{\frac{dy}{dx}}=\frac{dx}{dy}$ (Since $x=f^{-1}\left(y\right)$)

Hence proved.