# How do you differentiate f(x)=e^(-x)sinx using the product rule?

Dec 14, 2015

$f ' \left(x\right) = {e}^{- x} \left(\cos x - \sin x\right)$

#### Explanation:

The product rule is as follows:
if your function is a product of two functions,

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

then the derivative is

$f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$

In your case, $g \left(x\right) = {e}^{- x}$ and $h \left(x\right) = \sin x$.

Let's find the derivatives $g ' \left(x\right)$ and $h ' \left(x\right)$:

$g \left(x\right) = {e}^{- x} \textcolor{w h i t e}{\times x} \implies \textcolor{w h i t e}{x} g ' \left(x\right) = - {e}^{- x}$

$h \left(x\right) = \sin x \textcolor{w h i t e}{\times i} \implies \textcolor{w h i t e}{x} h ' \left(x\right) = \cos x$

So, in total, your derivative is:

$f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$

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

$\textcolor{w h i t e}{\times \times} = \textcolor{w h i t e}{x} {e}^{- x} \left(\cos x - \sin x\right)$