Question

Differentiate the given functions w.r.t. x.

$y=x^{sinx}+sin{x}^{cosx}$

Answer 1

Ley y = $x^{sinx}+sin{x}^{cosx}$

$Letu=x^{sinx},v=sin{x}^{cosx}$

$\therefore$ y = u + v

Differentiating w.r.t. x

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ........(i)

Now, $u=x^{sinx}$

Taking log on both sides, log u=(sin x) log x

Differentiating w.r.t. x,

$\frac{d}{dx}\left(logu\right)=sinx\frac{d}{dx}\left(logx\right)+logx\frac{d}{dx}\left(sinx\right)(\therefore Usingtheproductrule)\frac{1}{u}\frac{du}{dx}=[sinx\times \frac{1}{x}+logxcosx]\Rightarrow \frac{du}{dx}=u[\frac{sinx}{x}+cosxlogx]\Rightarrow \frac{du}{dx}=x^{sinx}[\frac{sinx}{x}+cosxlogx]Now,v=sin{x}^{cosx}Takinglogonbothsides,logv=cosxlog\left(sinx\right)Differentiatingw.r.t.x,\frac{d}{dx}\left(logv\right)=cosx\frac{d}{dx}log\left(sinx\right)+logsinx\frac{d}{dx}cosx\Rightarrow \frac{1}{v}\frac{dv}{dx}=[cosx\times \frac{1}{sinx}\times cosx+logsinx(-sinx\left)\right]\Rightarrow \frac{dv}{dx}=v[cotxcosx-sinxlog(sinx\left)\right]\Rightarrow \frac{dv}{dx}=sin{x}^{cosx}[cotxcosx-sinxlog(sinx\left)\right]Now,puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.\left(i\right)\frac{dy}{dx}=x^{sinx}[\frac{sinx}{x}+cosxlogx]+sin{x}^{cosx}[cotxcosx-sinxlog(sinx\left)\right]$