# How do you differentiate e^(4x)*sin(4x)?

Jul 19, 2017

d/(dx)[e^(4x)sinx] = color(blue)(e^(4x)cosx + 4e^(4x)sinx

#### Explanation:

We're asked to find the derivative

$\frac{d}{\mathrm{dx}} \left[{e}^{4 x} \sin x\right]$

We can first use the product rule, which states

$\frac{d}{\mathrm{dx}} \left[u v\right] = v \frac{\mathrm{du}}{\mathrm{dx}} u \frac{\mathrm{dv}}{\mathrm{dx}}$

where

• $u = {e}^{4 x}$

• $v = \sin x$:

$= {e}^{4 x} \frac{d}{\mathrm{dx}} \left[\sin x\right] + \sin x \frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right]$

The derivative of $\sin x$ is $\cos x$:

$= {e}^{4 x} \cos x + \sin x \frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right]$

Use the chain rule to differentiate the ${e}^{4 x}$ term:

$\frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] = \frac{d}{\mathrm{du}} \left[{e}^{u}\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = 4 x$

• $\frac{d}{\mathrm{du}} \left[{e}^{u}\right] = {e}^{u}$:

$= {e}^{4 x} \cos x + \left(\sin x\right) \left({e}^{4 x}\right) \frac{d}{\mathrm{dx}} \left[4 x\right]$

$= {e}^{4 x} \cos x + \left(\sin x\right) \left({e}^{4 x}\right) \left(4\right)$

or

= color(blue)(e^(4x)cosx + 4e^(4x)sinx