# How do you differentiate  f(x)=x^3e^xsin(x) using the product rule?

Dec 1, 2015

$f ' \left(x\right) = {x}^{2} {e}^{x} \left(3 \sin \left(x\right) + x \sin \left(x\right) + x \cos \left(x\right)\right)$

#### Explanation:

According to the product rule:

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right) h \left(x\right)\right] = f ' \left(x\right) g \left(x\right) h \left(x\right) + f \left(x\right) g ' \left(x\right) h \left(x\right) + f \left(x\right) g \left(x\right) h ' \left(x\right)$

$f ' \left(x\right) = {e}^{x} \sin \left(x\right) \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + {x}^{3} \sin \left(x\right) \frac{d}{\mathrm{dx}} \left[{e}^{x}\right] + {x}^{3} {e}^{x} \frac{d}{\mathrm{dx}} \left[\sin \left(x\right)\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$

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

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

Plug back in.

$f ' \left(x\right) = 3 {x}^{2} {e}^{x} \sin \left(x\right) + {x}^{3} {e}^{x} \sin \left(x\right) + {x}^{3} {e}^{x} \cos \left(x\right)$

$f ' \left(x\right) = {x}^{2} {e}^{x} \left(3 \sin \left(x\right) + x \sin \left(x\right) + x \cos \left(x\right)\right)$