# How do you differentiate f(x)=e^x*sin^2x using the product rule?

Aug 16, 2017

color(blue)(f'(x) = e^xsinx(sinx + 2cosx)

#### Explanation:

We're asked to find the derivative

$\frac{d}{\mathrm{dx}} \left[{e}^{x} \cdot {\sin}^{2} x\right]$

Using the power rule*, which is

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

where

• $u = {e}^{x}$

• $v = {\sin}^{2} x$:

$f ' \left(x\right) = {\sin}^{2} x \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] + {e}^{x} \frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right]$

The derivative of ${e}^{x}$ is ${e}^{x}$:

$f ' \left(x\right) = {e}^{x} {\sin}^{2} x + {e}^{x} \frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right]$

To differentiate the ${\sin}^{2} x$ term, we'll use the chain rule:

$\frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right] = \frac{d}{\mathrm{du}} \left[{u}^{2}\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = \sin x$

• $\frac{d}{\mathrm{du}} \left[{u}^{2}\right] = 2 u$ (from power rule):

$f ' \left(x\right) = {e}^{x} {\sin}^{2} x + {e}^{x} \cdot 2 \sin x \frac{d}{\mathrm{dx}} \left[\sin x\right]$

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

$f ' \left(x\right) = {e}^{x} {\sin}^{2} x + 2 {e}^{x} \sin x \cos x$

Or

color(blue)(ulbar(|stackrel(" ")(" "f'(x) = e^xsinx(sinx + 2cosx)" ")|)