ddx(e2xlogx)=e2xddxlogx+logx.ddxe2x=e2x.1x+logx.2.e2x=e2x(1x+2logx)

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

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$2 e^{2 x} \log x + \frac{e^{2 x}}{x}$

$e^{2 x} \log x + \frac{e^{2 x}}{x}$

$e^{x} \log x + \frac{e^{2 x}}{x}$

$e^{2 x} (\log x + x)$

answer is A.

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Detailed Solution

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

$= e^{2 x} \frac{d}{d x} \log x + \log x . \frac{d}{d x} e^{2 x}$

$= e^{2 x} . \frac{1}{x} + \log x .2. e^{2 x}$

$= e^{2 x} (\frac{1}{x} + 2 \log x)$

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