${(\sin x)^x} + {\sin ^{ - 1}}\sqrt x$

Let $y = {(\sin x)^x} + {\sin ^{ - 1}}\sqrt x = u + v,$ where

$u = {(\sin x)^x},v = {\sin ^{ - 1}}\sqrt x$

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

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

By taking log on both sides , we get

$\log u = x\log \sin x$

Differentating w.r.t. x, we get

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

$= x \cdot \cfrac{1}{{\sin x}} \cdot \cos x + \log \sin x = x\cot x + \log \sin x$

therefore, $\cfrac{{du}}{{dx}} = {(\sin x)^x}[x\cot x + \log \sin x]$ …..(ii)

Also, $v = {\sin ^{ - 1}}\sqrt x$
$\Rightarrow$ $\cfrac{{dv}}{{dx}} = \cfrac{1}{{\sqrt {1 - x} }} \cdot \cfrac{1}{{2\sqrt x }}$ …(iii)

From (i), (ii) \& (iii), we get

therefore, $\cfrac{{dy}}{{dx}} = {(\sin x)^x}[x \cdot \cot x + \log \sin x] + \cfrac{1}{{\sqrt {1 - x} }} \cdot \cfrac{1}{{2\sqrt x }}$

$= {(\sin x)^x}[x\cot x + \log \sin x] + \cfrac{1}{{2\sqrt {x - {x^2}} }}$