# How do you differentiate g(x) =e^xsinx using the product rule?

Jun 10, 2016

$g ' \left(x\right) = {e}^{x} \sin x + {e}^{x} \cos x$

#### Explanation:

The product rule tells us that $\frac{d}{\mathrm{dx}} \left(\textcolor{red}{u} \cdot \textcolor{b l u e}{v}\right) = \textcolor{red}{\frac{\mathrm{du}}{\mathrm{dx}}} \cdot \textcolor{b l u e}{v} + \textcolor{red}{u} \textcolor{b l u e}{\frac{\mathrm{dv}}{\mathrm{dx}}}$

In this question,
$\textcolor{red}{u} = \textcolor{red}{{e}^{x}}$, so $\textcolor{red}{\frac{\mathrm{du}}{\mathrm{dx}}} = \textcolor{red}{{e}^{x}}$,

and

$\textcolor{b l u e}{v} = \textcolor{b l u e}{\sin x}$, so $\textcolor{b l u e}{\frac{\mathrm{dv}}{\mathrm{dx}}} = \textcolor{b l u e}{\cos x}$.

$g ' \left(x\right) = \textcolor{red}{{e}^{x}} \cdot \textcolor{b l u e}{\sin x} + \textcolor{red}{{e}^{x}} \textcolor{b l u e}{\cos x}$

$= {e}^{x} \sin x + {e}^{x} \cos x$

Factor the ${e}^{x}$ if desired.