# How do you differentiate y=(5x^4-3x^2-1)(-5x^2+3) using the product rule?

Jun 26, 2017

$y ' = - 150 {x}^{5} + 120 {x}^{3} - 8 x$

#### Explanation:

Take two functions.
E.g. $f \left(x\right)$ and $g \left(x\right)$.
The product of those functions is:
$f \left(x\right) \cdot g \left(x\right) = \left(f \cdot g\right) \left(x\right)$

When we talk about derivatives, the product rule can be used to write the derivative of said functions. Using the functions from before, it states:
$\left(f \cdot g\right) ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

In your formula, we can see the "$\cdot$" symbol is between the brackets. So we could say something like:
$\left(5 {x}^{4} - 3 {x}^{2} - 1\right) \cdot \left(- 5 {x}^{2} + 3\right) = f \left(x\right) \cdot g \left(x\right)$.
$f \left(x\right) = 5 {x}^{4} - 3 {x}^{2} - 1$
$g \left(x\right) = - 5 {x}^{2} + 3$

$\left(f \cdot g\right) ' \left(x\right) = \left(5 {x}^{4} - 3 {x}^{2} - 1\right) ' \cdot \left(- 5 {x}^{2} + 3\right) + \left(5 {x}^{4} - 3 {x}^{2} - 1\right) \cdot \left(- 5 {x}^{2} + 3\right) '$
[Here we use the fact that we calculate each part seperately
and leave the constant, e.i $\left(5 {x}^{2} - 5 x\right) ' = \left(5 {x}^{2}\right) ' - \left(5 x\right) ' = 5 \left({x}^{2}\right) ' - 5 \left(x\right) '$]

$= \left(20 {x}^{3} - 6 x\right) \cdot \left(- 5 {x}^{2} + 3\right) + \left(5 {x}^{4} - 3 {x}^{2} - 1\right) \cdot \left(- 10 x\right)$
$= - 100 {x}^{5} + 60 {x}^{3} + 30 {x}^{3} - 18 x - 50 {x}^{5} + 30 {x}^{3} + 10 x$
$= - 150 {x}^{5} + 120 {x}^{3} - 8 x$