Question

Differentiate the following questions w.r.t. x.

$sin\left(ta{n}^{-1}{e}^{-x}\right).$

Answer 1

Ley y =$sin\left(ta{n}^{-1}{e}^{-x}\right).$

Differentiate both sides w.r.t. x, we get

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left[sin\right(ta{n}^{-1}\left(e^{-x}\right)\left)\right]$

$=cos\left\{ta{n}^{-1}\right(e^{-x}\left)\right\}\frac{d}{dx}\left\{tan{t}^{-1}\right(e^{-x}\left)\right\}\left(Usingchainrule\right)=cos\left\{tan{t}^{-1}\right(e^{-1}\left)\right\}\frac{1}{1+(e^{-x}{)}^2}\frac{d}{dx}\left(e^{-x}\right)=cos\left(tan{t}^{-1}\right(e^{-x}\left)\right)\frac{1}{1+e^{-2x}}.(-e^{-x})=-\frac{e^{-x}cos\left(ta{n}^{-1}{e}^{-x}\right)}{1+e^{-2x}}$