Respuesta :

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[tex]\text{Since (uv)'=u'v+v'u,}\\

\frac{d(\sqrt{x}\cdot e^x)}{dx}=\frac{d\sqrt{x}}{dx}e^x+ \frac{d e^x}{dx} \sqrt{x}\\

=\frac{1}{2\sqrt{x}} e^x + e^x \sqrt{x}[/tex]