# How do you differentiate g(x) = (x^2+1) (x^3-3) using the product rule?

Jan 25, 2016

$g ' \left(x\right) = \left(3 {x}^{4} + 3 {x}^{2}\right) + \left(2 {x}^{4} - 6 x\right)$

#### Explanation:

If we will consider

$u = \left({x}^{2} + 1\right)$
and
$v = \left({x}^{3} - 3\right)$

As product rule states:

$g ' \left(x\right) = u \cdot \frac{d}{\mathrm{dx}} v + v \cdot \frac{d}{\mathrm{dx}} u$

What this means is that the first function remains constant and multiplied into the derivative. Then we sum it up with the second function that is constant multiplied into the derivative of first function.

Hence,

$g ' \left(x\right) = \left({x}^{2} + 1\right) \frac{d}{\mathrm{dx}} \left({x}^{3} - 3\right) + \left({x}^{3} - 3\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$

$g ' \left(x\right) = \left({x}^{2} + 1\right) \cdot 3 {x}^{2} + \left({x}^{3} - 3\right) \cdot 2 x$

Here, we have taken the derivative. The derivative of $\left({x}^{3} - 3\right)$ is $3 {x}^{2}$ by the power rule and as $3$ is a constant number, it becomes zero. The same goes for the second function, where $1$ is a constant number and by power rule we get the derivative as $2 x$

$g ' \left(x\right) = \left(3 {x}^{4} + 3 {x}^{2}\right) + \left(2 {x}^{4} - 6 x\right)$

This is the answer for the given function.

Here we have multiplied the term that we got by differentiating inside the parentheses.

$3 {x}^{2} \cdot {x}^{2} = 3 {x}^{4}$
$3 {x}^{2} \cdot 1 = 3 {x}^{2}$
$2 x \cdot {x}^{3} = 2 {x}^{4}$
$2 x \cdot - 3 = - 6 x$