#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (ii) maths textbook solution

Answer: $\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}$

Hint:  Differentiate the equation taking log on both sides

Given: $y=x^{\sin x}+\sin ^{-1}(\sqrt{x})$

Solution:  $y=x^{\sin x}+\sin ^{-1}(\sqrt{x})$

$y=y_{1}+y_{2}$

Diff w.r.t x

$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$                    .......(1)

\begin{aligned} &\text { Let } y_{1}=x^{\sin x} \\\\ &\log y_{1}=\log x(\sin x)^{x} \end{aligned}

On diff both side with respect to x we get

$\frac{1}{y} \frac{d y}{d x}=\sin x \frac{d}{d x} \log x+\log x \frac{d}{d x} \sin x$

$\frac{1}{y} \frac{d y}{d x}=\sin x \frac{1}{x}+\log x \cos x$

$\frac{d y}{d x}=y\left(\frac{\sin x}{x}+\log x \cos x\right)$

$\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)$        from 1

$y_{2}=\sin ^{-1}(\sqrt{x})$

$\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \quad\left[\because \frac{d}{d x} \sin ^{-1}(\sqrt{x})=\frac{1}{\sqrt{1-x^{2}}}\right]$

\begin{aligned} &\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \\\\ &\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x(1-x)}} \end{aligned}

$\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$    ........(3)

Put (2) and (3) in eq(1)

$\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}$