Question

Find the derivative of given function
$f\left(x\right)={\cosh }^{-1}\left({\cosh }^{-1}x\right)$

Short answer

The answer of this question is$=\frac{1}{\left(\sqrt{x^2-1}\right)\sqrt{{\left({\cosh }^{-1}x\right)}^2-1}}$

Step-by-step solution

Step 1.Given information

We have been given

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

Step 2.Derivative of given function 

Derivative of product function

$\left(fg{)}^{\mathrm{\prime }}\right(x)=f^{\mathrm{\prime }}(x\left)g\right(x)+f(x\left)g^{\mathrm{\prime }}\right(x)$

Derivative of quotient function

${\left(\frac{f}{g}\right)}^{\mathrm{\prime }}\left(x\right)=\frac{f^{\mathrm{\prime }}\left(x\right)g\left(x\right)-f\left(x\right)g^{\mathrm{\prime }}\left(x\right)}{\left(g\right(x){)}^2}$

Chain rule of derivation

$(f\circ u{)}^{\mathrm{\prime }}(x)=f^{\mathrm{\prime }}(u\left(x\right)\left)u^{\mathrm{\prime }}\right(x)$

Solving we get

$\begin{array}{r}\frac{d}{dx}(\sinh x)=\cosh x\\ \frac{d}{dx}(\cosh x)=\sinh x\\ \frac{d}{dx}(\tanh x)={\mathrm{sech}}^{2}x\\ \frac{d}{dx}\left({\sinh }^{-1}x\right)=\frac{1}{\sqrt{x^2+1}}\\ \frac{d}{dx}\left({\cosh }^{-1}x\right)=\frac{1}{\sqrt{x^2-1}}\\ \frac{d}{dx}\left({\tanh }^{-1}x\right)=\frac{1}{1-x^2}\end{array}$

Step 3. Using chain rule 

$\begin{array}{r}\frac{d}{dx}{\cosh }^{-1}\left({\cosh }^{-1}x\right)=\frac{1}{\sqrt{{\left({\cosh }^{-1}x\right)}^2-1}}\left(\frac{d}{dx}{\cosh }^{-1}x\right)\text{[Using chain rule]}\\ =\frac{1}{\sqrt{{\left({\cosh }^{-1}x\right)}^2-1}\left(\frac{1}{\sqrt{x^2-1}}\right)}\\ =\frac{1}{\left(\sqrt{x^2-1}\right)\sqrt{{\left({\cosh }^{-1}x\right)}^2-1}}\\ =\frac{1}{\left(\sqrt{x^2-1}\right)\sqrt{{\left({\cosh }^{-1}x\right)}^2-1}}\end{array}$