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Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 44 maths textbook solution

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Answer:  2 e^{x} \cot 2 x+e^{x} \log \sin 2 x

Hint: you must know about the rules of solving derivative of exponential logarithm and trigon function

Given: e^{x} \log \sin 2 x


Let  y=e^{x} \log \sin 2 x

Differentiate with respect to x

\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[e^{x} \log \sin 2 x\right]

\begin{aligned} &=e^{x} \frac{\mathrm{d}}{\mathrm{dx}}(\log \sin 2 x)+(\log \sin 2 x) \frac{\mathrm{d}}{\mathrm{dx}}\left(e^{x}\right) \\\\ &=e^{x} \frac{1}{\sin 2 x} \frac{\mathrm{d}}{\mathrm{dx}} \sin 2 x+\log \sin 2 x\left(e^{x}\right) \end{aligned}

\begin{aligned} &=\frac{e^{x}}{\sin 2 x} \cos 2 x \frac{\mathrm{d}}{\mathrm{dx}}(2 x)+e^{x} \log \sin 2 x \\\\ &=\frac{2 \cos 2 x e^{x}}{\sin 2 x}+e^{x} \log \sin 2 x \\\\ &\Rightarrow 2 e^{x} \cot 2 x+e^{x} \log \sin 2 x \end{aligned}

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