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Find $\frac{d}{dx} {\Large(} \sqrt{x}+6e^x {\Large)}$.

A

$\cfrac{\sqrt{x}}{2}+6e^x$

B

$\cfrac{2}{\sqrt{x}}+6xe^{x-1}$

C

$\cfrac{1}{2\sqrt{x}}+6xe^{x-1}$

D

$\cfrac{1}{2\sqrt{x}}+6e^x$

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