Question

If $y=f\left(x\right)$ is a derivable function of $x$ such that the inverse function $x=f^{-1}\left(y\right)$ is defined.
then show that $\frac{dx}{dy}=\frac{1}{\left(\frac{dy}{dx}\right)}1$ where $\frac{dy}{dx}\ne 0$

Answer 1

Let $\delta x$ and $\delta y$ be small change in $x$ and $y$ respectively.

As $\delta x\to 0,\delta y\to 0$

We can say that $\frac{\delta y}{\delta x}\times \frac{\delta x}{\delta y}=1$

Take $\underset{\delta x\to 0}{lim}$ on both sides.

$\therefore \underset{\delta x\to 0}{lim}(\frac{\delta y}{\delta x}\times \frac{\delta x}{\delta y})=\underset{\delta x\to 0}{lim}1$

$\Rightarrow \underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}\times \underset{\delta x\to 0}{lim}\frac{\delta x}{\delta y}=1$

Since, $\delta y\to 0$ as $\delta x\to 0$,

$\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}\times \underset{\delta y\to 0}{lim}\frac{\delta x}{\delta y}=1$

We know that $\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}=\frac{dy}{dx}$ and $li{m}_{\delta y\to 0}\frac{\delta x}{\delta y}=\frac{dx}{dy}$

$\therefore \frac{dy}{dx}\times \frac{dx}{dy}=1$

$\therefore \frac{dx}{dy}=\frac{1}{\left(\frac{dy}{dx}\right)}$