Question

If $y=x^{sinx}+sin\left(x^x\right),$ Find $\frac{dy}{dx}$

Answer 1

$y=x^{sinx}+sin\left(x^x\right)$

Let $u=x^{sinx}$

And, $v=sin\left(x^x\right)$

Now,

$u=x^{sinx}$

$logu=sinxlogx$

$\frac{1}{u}\frac{du}{dx}=\frac{sinx}{x}+logxcosx$

$\frac{du}{dx}=x^{sinx}[\frac{sinx}{x}+logxcosx]$

Now,

$v=sin\left(x^x\right)$

$\frac{dv}{dx}=cos\left(x^x\right)\frac{d}{dx}\left(x^x\right)$

Now, let $w=x^x$

$logw=log{x}^x$

$logw=xlogx$

$=\frac{1}{w}\frac{dw}{dx}=1+logx$

$\frac{dw}{dx}=x^x(1+logx)$

Therefore,

$\frac{dv}{dx}=x^{x}cos\left(x^x\right)(1+logx)$

Hence,

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

$\frac{dy}{dx}=x^{sinx}[\frac{sinx}{x}+logxcosx]+x^{x}cos\left(x^x\right)(1+logx)$