Question

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

$cos(logx+e^x).$

Answer 1

Ley y = $cos(logx+e^x).$

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

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left\{cos\right(logx+e^x\left)\right\}\Rightarrow \frac{dy}{dx}=-sin(logx+e^x)\frac{d}{dx}(logx+e^x)\left(Usingchainrule\right)\Rightarrow \frac{dy}{dx}=-sin(logx+e^x)(\frac{1}{x}+e^x)\Rightarrow \frac{dy}{dx}=\frac{-(x{e}^x+1)sin(logx+e^x)}{x}$