# How do you use the Product Rule to find the derivative of (x^2)(x^3+4)?

Dec 25, 2015

#### Answer:

The explanation is given below.

#### Explanation:

Product rule for differentiation where product of functions is given.

If $f \left(x\right)$ and $g \left(x\right)$ are two functions then their product would be $f \left(x\right) \cdot g \left(x\right)$

The derivative $\left(f \left(x\right) \cdot g \left(x\right)\right) '$ is given by the product rule

$\left(f \left(x\right) g \left(x\right)\right) ' = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$

Derivative rules

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) + g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$
$\frac{d}{\mathrm{dx}} \left(c\right) = 0$ Derivative of constant is zero.

Now coming to our problem

$y = \left({x}^{2}\right) \left({x}^{3} + 4\right)$

$f \left(x\right) = {x}^{2}$ and $g \left(x\right) = \left({x}^{3} + 4\right)$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right) \implies f ' \left(x\right) = 2 x$
$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} + 4\right) = \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(4\right)$
$g ' \left(x\right) = 3 {x}^{2} + 0$
$g ' \left(x\right) = 3 {x}^{2}$

Using product rule

$\frac{d}{\mathrm{dx}} \left(\left({x}^{2}\right) \left({x}^{3} + 4\right)\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({x}^{3} + 4\right) + \left({x}^{3} + 4\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$= {x}^{2} \left(3 {x}^{2}\right) + \left({x}^{3} + 4\right) \left(2 x\right)$

$= 3 {x}^{4} + 2 {x}^{4} + 8 x$ simplifying

$= 5 {x}^{4} + 8 x$ Answer