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#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 25 maths textbook solution

Answer: $\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{x}(\log \sin x)+x \cdot \cot x$

Hint: Differentiate the equation taking log on both sides

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

Solution:

\begin{aligned} &\text { Let } u=x^{\sin x} \text { and } v=(\sin x)^{x} \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\\\ &u=x^{\sin x} \end{aligned}

\begin{aligned} &\log u=\log x^{\sin x} \\\\ &\frac{d}{d x}(\log u)=\frac{d}{d x}(\sin x \cdot \log x) \end{aligned}

$\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(\sin x) \cdot \log x+\sin x \frac{d}{d x}(\log x) \quad\left[\because \frac{d}{d x}(\log x)=\frac{1}{x}\right]$

$\frac{d u}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right] \quad\left[\because \frac{d}{d x}(\sin x)=\cos x\right]$

\begin{aligned} &v=(\sin x)^{x} \\\\ &\log v=\log (\sin x)^{x} \\\\ &\frac{d}{d x} v=\frac{d}{d x}[x \log (\sin x)] \end{aligned}

$\frac{1}{v} \frac{d v}{d x}=\cdot \log (\sin x)+x \cdot \frac{d}{d x}(\log \sin x)$

$=\log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)$

\begin{aligned} &\frac{d v}{d x}=(\sin x)^{x}[\log \sin x+x \cdot \cot x] \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \end{aligned}

$\frac{d y}{d x}=x^{\sin x}\left[\cos x \log x+\frac{\sin x}{x}\right]+(\sin x)^{x}(\log \sin x)+x \cdot \cot x$