Question

Find the derivative of$\left[\frac{{\tan }^{-1}x}{(1+{\tan }^{-1}x)}\right]$ with respect to ${\tan }^{-1}x$.

Answer 1

Find the required derivative.

Given function : $\left[\frac{{\tan }^{-1}x}{(1+{\tan }^{-1}x)}\right]$

The derivative of $\left[\frac{{\tan }^{-1}x}{(1+{\tan }^{-1}x)}\right]$ with respect to ${\tan }^{-1}x$ is

$$\begin{gathered} \frac{d}{d\tan x^{-1}x}\left[\frac{{\tan }^{-1}x}{(1+{\tan }^{-1}x)}\right]=\frac{1\times (1+{\tan }^{-1}x)-\left({\tan }^{-1}x\right)\times 1}{{(1+{\tan }^{-1}x)}^2}\left[\because f\text{'}\left(x\right)=\frac{u\text{'}\left(x\right)\times v\left(x\right)-u\left(x\right)\times v\text{'}\left(x\right)}{\left[v{\left(x\right)}^2\right]}\right] \\ =\frac{1+{\tan }^{-1}x-{\tan }^{-1}x}{{(1+{\tan }^{-1}x)}^2} \\ =\frac{1}{{(1+{\tan }^{-1}x)}^2} \end{gathered}$$

Hence, the derivative of$\left[\frac{{\tan }^{-1}x}{(1+{\tan }^{-1}x)}\right]$ with respect to ${\tan }^{-1}x$ is $\frac{1}{{(1+{\tan }^{-1}x)}^2}$.