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#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 28 sub question (i) maths

Answer:  $\frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$

Hint: Differentiate the equation taking log on both sides

Given:

Solution:

\begin{aligned} &y=(\sin x)^{x}+\sin ^{-1}(\sqrt{x}) \\\\ &y=y_{1}+y_{2} \end{aligned}

Diff w.r.t x

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

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

Taking log on both sides

$\log y_{1}=\log (\sin x)^{x} \quad\left[\because \log m^{n}=n \log m\right]$

$\log y_{1}=x \log \sin x$

$\frac{1}{y_{1}} \frac{d y}{d x}=x \cdot \frac{1}{\sin x} \cdot \cos x+\log \sin x \cdot 1$                        $\left[\begin{array}{l} \because \frac{d}{d x} \log x=\frac{1}{x} \\\\ \frac{d}{d x} I . I I=I \frac{d}{d x} I I+I I \frac{d}{d x} I \end{array}\right]$

\begin{aligned} &\frac{d y_{1}}{d x}=y_{1}[x \cdot \cot x+\log \sin x] \\\\ &\frac{d y_{1}}{d x}=(\sin x)^{n}[x . \cot x+\log \sin x] \end{aligned}            .......(2)

\begin{aligned} &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}) \end{aligned}

\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)

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