Question

Fill in the blank:

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

Short answer

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

Step-by-step solution

Given Information 

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

Simplification 

We know that, $\cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right).$

we compute the derivative of $\cos h^{-1}x.$

We get,

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