A) \[{{e}^{x}}(\log \sin 2x+2\cot 2x)\]
B) \[{{e}^{x}}(\log \cos 2x+2\cot 2x)\]
C) \[{{e}^{x}}(\log \cos 2x+\cot 2x)\]
D) None of these
Correct Answer: A
Solution :
\[\frac{d}{dx}({{e}^{x}}\log \sin 2x)={{e}^{x}}\log \sin 2x+2{{e}^{x}}\frac{1}{\sin 2x}\cos 2x\] \[={{e}^{x}}\log \sin 2x+{{e}^{x}}2\cot 2x\]\[={{e}^{x}}(\log \sin 2x+2\cot 2x).\]
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