${x^{\sin x}} + {(\sin x)^{\cos x}}$

Let$y = {x^{\sin x}} + {(\sin x)^{\cos x}} = u + v$ , where

$u = {x^{\sin x}}and\;v = {(\sin x)^{\cos x}}$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$

….(i)

Now, $u = {x^{\sin x}}$

By taking log on both sides , we get

$\log u = \sin x\log x$ ….(ii)

Differentiating (ii) w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{u}\cfrac{{du}}{{dx}} = \sin x\cfrac{d}{{dx}}(\log x) + \log x\cfrac{d}{{dx}}(\sin x)$
$= \sin x \cdot \cfrac{1}{x} + \log x\cos x$

therefore, $\cfrac{{du}}{{dx}} = {x^{\sin x}}\left[ {\cfrac{{\sin x}}{x} + \log x\cos x} \right]$ …(iii)

Also, $v = {(\sin x)^{\cos x}}$

By taking log on both sides , we get

$\log v = \cos x\log \sin x$ …(iv)

Differentiating (iv) w.r.t. x, we get

$\cfrac{1}{v}\cfrac{{dv}}{{dx}} = \cos x\cfrac{d}{{dx}}\log \sin x + \log \sin x\cfrac{d}{{dx}}\cos x$

$= \cos x\cfrac{1}{{\sin x}}\cos x + \log \sin x( - \sin x)$

$= \cos x\cot x - \sin x\log \sin x$

therefore, $\cfrac{{dv}}{{dx}} = {(\sin x)^{\cos x}}[\cos x\cot x - \sin x\log \sin x]$ ….(v)

Substituting the values of (iii) \& (v) in (i), we get

$\cfrac{{dy}}{{dx}} = {x^{\sin x}}\left[ {\cfrac{{\sin x}}{x} + \log x\cos x} \right] + {(\sin x)^{\cos x}}[\cos x\cot x - \sin x\log \sin x]$