Question

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

Answer 1

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

Let

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

Now,

$u={\left(\sin x\right)}^x$

Taking $\log$ both sides, we have

$\log u=x\log \left(\sin x\right)$

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

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

$\frac{du}{dx}={\left(\sin x\right)}^x(\log \left(\sin x\right)+x\cot x)$

Similarly,

$v=x^{\sin x}$

Taking $\log$ both sides, we get

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

Differentiating above equation w.r.t. $x$, we have

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

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

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

Therefore,

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

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