Question

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

Answer 1

Let $u=x^{\sin x}$ and $v={\sin x}^{\cos x}$

$\Rightarrow y=u+v$

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

Consider $u=x^{\sin x}$

$\log u=\sin x\log x$

$\frac{1}{u}\frac{du}{dx}=\frac{\sin x}{x}+\log x\left(\cos x\right)$

$\Rightarrow \frac{du}{dx}=u(\frac{\sin x}{x}+\cos x\log x)$

$\Rightarrow \frac{du}{dx}=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)$

Consider $v={\left(\sin x\right)}^{\cos x}$

$\log v=\cos x\log \sin x$

$\frac{1}{v}\frac{dv}{dx}=\frac{\cos x}{\sin x}\left(\cos x\right)+\log \sin x(-\sin x)$

$\Rightarrow \frac{dv}{dx}={\left(\sin x\right)}^{\cos x}(\frac{{\cos }^{2}x}{\sin x}-\sin x\log \sin x)$

$\therefore \frac{dy}{dx}=x^{\sin x}(\frac{\sin x}{x}+\cos x\log x)+{\left(\sin x\right)}^{\cos x}(\frac{{\cos }^{2}x}{\sin x}-\sin x\log \sin x)$