frac{d}{dx}(e^x log sin 2x) = | vedclass

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(A) To find the derivative of $e^x \log \sin 2x$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = e^x$ and $v

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$e^x(\log \sin 2x + 2\cot 2x)$

Vedclass · Answer

(A) To find the derivative of $e^x \log \sin 2x$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = e^x$ and $v = \log \sin 2x$.Then $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = \frac{1}{\sin 2x} \cdot \frac{d}{dx}(\sin 2x) = \frac{1}{\sin 2x} \cdot (2 \cos 2x) = 2 \cot 2x$.Applying the product rule:$\frac{d}{dx}(e^x \log \sin 2x) = e^x \cdot (2 \cot 2x) + (\log \sin 2x) \cdot e^x$.Factoring out $e^x$,we get:$= e^x(\log \sin 2x + 2 \cot 2x)$.

$\frac{d}{dx}(e^x \log \sin 2x) = $

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