Question

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

• A. cosec hx
• B. sec hx
• C. sin hx
• D. cos hx

Answer 1

The correct option is B sec hx

Given,

$\frac{d}{dx}\left({\tan }^{-1}\left(sinhx\right)\right)$

let $f={\tan }^{-1}\left(u\right),u=\sinh \left(x\right)$

$=\frac{d}{du}\left({\tan }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\sinh \left(x\right)\right)$

$=\frac{1}{u^2+1}\cosh \left(x\right)$

$=\frac{1}{{\sinh }^2\left(x\right)+1}\cosh \left(x\right)$

$=\frac{\cosh \left(x\right)}{{\cosh }^2\left(x\right)}$

$=\frac{1}{\cosh \left(x\right)}$

$=\sec hx$