608.

Differentiate the following function w.r.t. x:

y = ${\left(\mathrm{sinx}\right)}^{\mathrm{x}} + {\sin }^{-1}\sqrt{\mathrm{x}}$

$$\begin{gathered} \mathrm{Y} = {\left( \sin x \right)}^{\mathrm{x}} + {\sin }^{-1} \sqrt{\mathrm{x}} \\ \mathrm{Let} \mathrm{u} = {\left( \sin x \right)}^{\mathrm{x}} \mathrm{and} \mathrm{v} = {\sin }^{-1} \sqrt{\mathrm{x}} \end{gathered}$$

Now y = u + v

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} + \frac{\mathrm{dv}}{\mathrm{dx}} ......\left(\mathrm{i}\right) \\ \mathrm{Consider} \mathrm{u} = {\left( \sin \mathrm{x} \right)}^{\mathrm{x}} \end{gathered}$$

Taking logarithms on both sides, we have,

log u = x log ( sin x )

Differentiating with respect to x, we have,

$$\begin{gathered} \frac{1}{\mathrm{u}}. \frac{\mathrm{du}}{\mathrm{dx}} = \log \left( \sin \mathrm{x} \right) + \frac{\mathrm{x}}{\sin {\displaystyle \mathrm{x}}}. \cos \mathrm{x} \\ \Rightarrow \frac{{\displaystyle \mathrm{du}}}{{\displaystyle \mathrm{dx}}} = {\left( \sin \mathrm{x} \right)}^{\mathrm{x}} \left(\log \left( \sin \mathrm{x} \right) + \mathrm{x} \cot \mathrm{x} \right) ....\left(\mathrm{ii}\right) \\ \mathrm{Consider} \mathrm{v} = {\sin }^{-1}\sqrt{\mathrm{x}} \\ \frac{{\displaystyle \mathrm{dv}}}{{\displaystyle \mathrm{dx}}} = \frac{1}{\sqrt{1 - \mathrm{x}}} \mathrm{x} \frac{1}{2\sqrt{\mathrm{x}}} ....\left(\mathrm{iii}\right) \end{gathered}$$

From (i), (ii)  and (iii)

We get, $\frac{\mathrm{dy}}{\mathrm{dx}} = {\left( \sin \mathrm{x} \right)}^{\mathrm{x}} \left(\log \left(\sin \mathrm{x}\right) + \mathrm{x} \cot \mathrm{x} \right) + \frac{1}{2\sqrt{\mathrm{x}} . \sqrt{1 - \mathrm{x}}}$