Question

If $y=(\sin x{)}^x+x^x$ then find $\frac{dy}{dx}$

Answer 1

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

Let $u=x^x$ and $v={\left(sinx\right)}^x$

Therefore $y=u+v$

Differentiating above equation,

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

Since $u=x^x$

Taking log on both sides, $logu=xlogx$

Differentiating both,$\frac{1}{u}\frac{dy}{dx}=x\frac{1}{x}+logx$

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