# How do you find the derivative of y=e^x(sinx+cosx)?

Feb 25, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {e}^{x} \cos x$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{product rule}}$

$\text{Given "y=g(x).h(x)" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } g \left(x\right) = {e}^{x} \Rightarrow g ' \left(x\right) = {e}^{x}$

$\text{and } h \left(x\right) = \sin x + \cos x \Rightarrow g ' \left(x\right) = \cos x - \sin x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left(\cos x - \sin x\right) + {e}^{x} \left(\sin x + \cos x\right)$

$\textcolor{w h i t e}{\times \times x} = {e}^{x} \cos x \cancel{- {e}^{x} \sin x} \cancel{{e}^{x} \sin x} + {e}^{x} \cos x$

$\textcolor{w h i t e}{\times \times x} = 2 {e}^{x} \cos x$