Question

Fill in the blank:

$\frac{d}{dx}\left(\tan h^{-1}x\right)=\_\_\_\_\_\_\_\_\_\_\_.$

Short answer

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

Step-by-step solution

Given Information  

The given expression is $\frac{d}{dx}\left(\tan h^{-1}x\right).$

Simplification  

We know that,

$\tan h^{-1}x=\frac{1}{2}\{\ln \left(x+1\right)-\ln \left(1-x\right)\}.$

we compute the derivative of $\frac{d}{dx}\left(\tan h^{-1}x\right).$

We get,

$$\begin{gathered} \frac{d}{dx}\left(\tan h^{-1}x\right) \\ =\frac{d}{dx}\left[\frac{1}{2}\right\{\ln \left(x+1\right)-\ln \left(1-x\right)\} \\ =\frac{1}{2}\{\frac{1}{(x+1)}.1-\frac{1}{(1-x)}.-1\} \\ =\frac{1}{2}(\frac{1}{x+1}+\frac{1}{1-x}) \\ =\frac{1}{2}.\frac{2}{(1-x^2)} \\ =\frac{1}{1-x^2} \\ Therefore,\frac{d}{dx}(\tan h^{-1}x)=\frac{1}{1-x^2}. \end{gathered}$$