Question

Differentiate the function with respect to $x$:
$x^{sin\,x}+(sin\,x)^{cos\,x}$

Answer 1

The correct answer is $\frac{dy}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}]+(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx]$
Let $y=x^{sin\,x}+(sin\,x)^{cosx}$
Also,let $u=x^{sinx}$ and $v=(sinx)^{cosx}$
$∴y=u+v$
$⇒\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} ....(1)$
$u=x^{sinx}$
$⇒log\,u=log(x^{sin\,x})$
$⇒log\,u=sin\,x\,log\,x$
Differentiating both sides with respect to $x$,we obtain
$⇒\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(sin\,x)log(x)+sinx.\frac{d}{dx}[log\,x]$
$⇒\frac{du}{dx}=u[cos\,x\,log\,x+sinx.\frac{1}{x}]$
$⇒\frac{du}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}] ..(2)$
$v=(sin\,x)^{cosx}$
$⇒log\,v=log(sin\,x)^{cosx}$
$⇒log\,v=cos\,x\,log(sin\,x)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(cosx)\times log(sinx)+cosx\frac{d}{dx}(log(sinx))$
$⇒\frac{dv}{dx}=v[-sin\,x\,log(sinx)+cosx.\frac{1}{sinx}.\frac{d}{dx}(sinx)]$
$⇒\frac{dv}{dx}=(sin\,x)^{cos\,x}[-sin\,x\,log\,sin\,x+\frac{cos\,x}{sin\,x}cos\,x]$
$⇒\frac{dv}{dx}=(sinx)^{cosx}[-sinx\,log\,sinx+cotx\,cosx]$
$⇒\frac{dv}{dx}=(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx] ...(3)$
Therefore,from (1),(2),and(3),we obtain
$\frac{dy}{dx}=x^{sin\,x}[cos\,x\,log\,x+\frac{sinx}{x}]+(sinx)^{cosx}[cotx\,cosx-sinx\,log\,sinx]$