Let $\delta y$ be the increment in $y$ corresponding to an increment $\delta x$ in $x$.
as $\delta x \rightarrow 0, \delta y \rightarrow 0$
Now y is a differentiable function of x.
$\therefore \lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x}$
Now $\frac{\delta y}{\delta x} \times \frac{\delta x}{\delta y}=1$
$\therefore \frac{\delta x}{\delta y}=\frac{1}{\frac{\delta y}{\delta x}}$
Taking limits on both sides as $\delta x \rightarrow 0, w e \geq t$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\lim _{\delta x \rightarrow 0}\left[\frac{1}{\frac{\delta y}{\delta x}}\right]=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}}$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}} \quad \ldots .[$ as $\delta x \rightarrow 0, \delta y \rightarrow 0]$
Since limit in R.H.S. exists
limit in L.H.S. also exists and we have,
$\lim _{\delta y \rightarrow 0} \frac{\delta x}{\delta y}=\frac{d x}{d y}$
$\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$, where $\frac{d y}{d x} \neq 0$
Let $y=\tan ^{-1} x$
$x=\tan y \Rightarrow \cos y=\frac{1}{\sqrt{1+\tan ^2 y}}=\frac{1}{\sqrt{1+x^2}}$
$\therefore \sec ^y \cdot \frac{d y}{d x}=1 \Rightarrow \frac{d x}{d y}=\sec ^2 y$
$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\sec ^2 y}=\cos ^2 y \Rightarrow \frac{d y}{d x}=\cos ^y$
$\frac{d\left(\tan ^{-1} x\right)}{d x}=\cos ^2 y=(\cos y)^2=\left(\frac{1}{\sqrt{1+x^2}}\right)^2$
$\therefore \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$
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