# How do you find the derivative of 2e^(-0.7x)sin(2pix)?

Mar 6, 2017

$= 2 {e}^{- 0.7 x} \left(- 0.7 \sin \left(2 \pi x\right) + 2 \pi \cos \left(2 \pi x\right)\right)$

#### Explanation:

You would apply the rules:

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right) \to f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$
$f \left(x\right) = {e}^{g} \left(x\right) \to f ' \left(x\right) = {e}^{g} \left(x\right) \cdot g ' \left(x\right)$
$f \left(x\right) = \sin \left(g \left(x\right)\right) \to f ' \left(x\right) = \cos \left(g \left(x\right)\right) \cdot g ' \left(x\right)$
$f \left(x\right) = k g \left(x\right) \to f ' \left(x\right) = k g ' \left(x\right)$

and get

$y ' = 2 {e}^{- 0.7 x} \cdot \left(- 0.7\right) \cdot \sin \left(2 \pi x\right) + 2 {e}^{- 0.7 x} \cdot \cos \left(2 \pi x\right) \cdot 2 \pi$

$= 2 {e}^{- 0.7 x} \left(- 0.7 \sin \left(2 \pi x\right) + 2 \pi \cos \left(2 \pi x\right)\right)$