Question

Use implicit differentiation and the fact that $\tan \left({\tan }^{-1}x\right)=x$for all $x$in the domain of ${\tan }^{-1}x$ to prove that $\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$ .

Short answer

We proved$\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$.

Step-by-step solution

Step 1. Given Information 

We are given ${\tan }^{-1}x$and using implicit differentiation we need to prove that$\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$.

Step 2. Proving the expression 

$\tan \left({\tan }^{-1}x\right)=x$

Differentiating both sides we get,

$$\begin{gathered} \frac{d}{dx}\left\{\tan \left({\tan }^{-1}x\right)\right\}=\frac{d}{dx}\left(x\right) \\ \Rightarrow se{c}^2\left({\tan }^{-1}x\right)\frac{d}{dx}\left({\tan }^{-1}x\right)=1 \\ \Rightarrow \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{se{c}^2\left({\tan }^{-1}x\right)} \\ \Rightarrow \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+{\tan }^2\left({\tan }^{-1}x\right)} \\ \Rightarrow \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2} \end{gathered}$$