Question

Differentiate the function w.r.t. $x$.
$x^{\sin x}+(\sin x{)}^{\cos x}$

Answer 1

Let $y=x^{\sin x}+(\sin x{)}^{\cos x}$
Also, let $u=x^{\sin x}$ and $v=(\sin x{)}^{\cos x}$
$\therefore y=u+v$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ...(1)
$u=x^{\sin x}$
$\Rightarrow \log u=\log \left(x^{\sin x}\right)$
$\Rightarrow \log u=\sin x\log x$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(\sin x\right).\log x+\sin x.\frac{d}{dx}\left(\log x\right)$
$\Rightarrow \frac{du}{dx}=u[\cos x\log x+\sin x.\frac{1}{x}]$
$\Rightarrow \frac{du}{dx}=x^{\sin x}[\cos x\log x+\frac{\sin x}{x}]$ ...(2)
$v=(\sin x{)}^{\cos x}$
$\log v=\log (\sin x{)}^{\cos x}$
$\log v=\cos x\log \left(\sin x\right)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left(\cos x\right)\times \log \left(\sin x\right)+\cos x\times \frac{d}{dx}\left[\log \right(\sin x\left)\right]$
$\Rightarrow \frac{dv}{dx}=v[-\sin x\log (\sin x)+\cos x.\frac{1}{\sin x}.\frac{d}{dx}(\sin x\left)\right]$
$\Rightarrow \frac{dv}{dx}=(\sin x{)}^{\cos x}[-\sin x\log \sin x+\frac{\cos x}{\sin x}\cos x]$
$\Rightarrow \frac{dv}{dx}=(\sin x{)}^{\cos x}[-\sin x\log \sin x+\cot x\cos x]$
$\Rightarrow \frac{dv}{dx}=(\sin x{)}^{\cos x}[\cot x\cos x-\sin x\log \sin x]$ ...(3)
From (1), (2) and (3), we obtain
$\frac{dy}{dx}=x^{\sin x}[\cos x\log x+\frac{\sin x}{x}]+(\sin x{)}^{\cos x}[\cot x\cos x-\sin x\log \sin x]$