Question

Solve : $\cos (\log x+e^x),x>0$

Answer 1

Let $y=\cos (\log x+e^x)$

differentiating both sides w.r.t $x$

$\frac{dy}{dx}=\frac{d\left(\cos \right(\log x+e^x\left)\right)}{dx}$

$=\frac{-\sin (\log x+e^x)}{dx}\frac{d(\log +e^x)}{dx}$

$(\because \frac{d\cos x}{dx}=-\sin x)$

$\frac{dy}{dx}=-\sin (\log x+e^x)(\frac{d\left(\log x\right)}{dx}+\frac{d{e}^x}{dx})$

$=-\sin (\log x+e^x)(\frac{1}{x}+e^x)$

$\frac{dy}{dx}=-(\frac{1}{x}+e^x)\left(\sin \right(\log x+e^x\left)\right)$