How do you find the derivative for y= e^-x sinx?

Mar 15, 2018

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

Explanation:

Given: $y = {e}^{- x} \sin x$

Use the product rule: $\left(u \cdot v\right) ' = u \cdot v ' + v \cdot u '$

Use the chain rule: $\left({e}^{u}\right) ' = u ' {e}^{u}$

Use $\left(\sin x\right) ' = \cos x$

Let u = e^(-x) ;" " v = sin x

u' = -e^(-x); " " v' = cos x

Using the product rule:

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

Rearrange: $\text{ } y ' = {e}^{- x} \cos x - {e}^{- x} \sin x$

Factor: $\text{ } y ' = {e}^{- x} \left(\cos x - \sin x\right)$