Question

Prove each of the differentiation formulas in Exercises 93–96. (These exercises involve hyperbolic functions.)

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

Short answer

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

Step-by-step solution

Step 1. Given Information  

We need to prove the hyperbolic function$\frac{d}{dx}\left(\tan h^{-1}x\right)=\frac{1}{1-x^2}$.

Step 2. Proving the function 

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