Question

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

Answer 1

`y' = -150x^5+120x^3-8x`

Explanation:

Take two functions.
E.g. `f(x)` and `g(x)`.
The product of those functions is:
`f(x)*g(x) = (f*g)(x)`

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:
`(f * g)'(x) = f'(x) * g(x) + f(x) * g'(x)`

In your formula, we can see the "`*`" symbol is between the brackets. So we could say something like:
`(5x^4-3x^2-1) * (-5x^2+3) = f(x) * g(x)`.
`f(x) = 5x^4-3x^2-1`
`g(x) = -5x^2+3`

`(f * g)'(x) = (5x^4-3x^2-1)' * (-5x^2+3) + (5x^4-3x^2-1) * (-5x^2+3)'`
[Here we use the fact that we calculate each part seperately
and leave the constant, e.i `(5x^2-5x)' = (5x^2)'-(5x)' = 5(x^2)' - 5(x)'`]

`= (20x^3-6x) * (-5x^2+3) + (5x^4-3x^2-1) * (-10x)`
`= -100x^5+60x^3+30x^3-18x -50x^5+30x^3+10x`
`= -150x^5+120x^3-8x`