Question

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

Answer 1

The answer is `(3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3)`, which simplifies to `10x^4+12x^3-3x^2-18x-15`.

Explanation:

According to the product rule,
`( f ⋅ g ) ′ = f ′ ⋅ g + f ⋅ g ′`
This just means that when you differentiate a product, you do derivative of the first, leave second alone, plus derivative of the second, leave the first alone.

So the first would be `(x^3 − 3x)` and the second would be `(2x^2 + 3x + 5)`.

Okay, now the derivative of the first is `3x^2-3`, times the second is `(3x^2-3)*(2x^2 + 3x + 5)`.

The derivative of the second is `(2*2x+3+0)`, or just `(4x+3)`.
Multiply it by the first and get `(x^3 − 3x)*(4x+3)`.

Add both portions together now: `(3x^2-3)*(2x^2 + 3x + 5) + (x^3 − 3x)*(4x+3)`

If you multiply it all out and simplify, you should get `10x^4+12x^3-3x^2-18x-15`.

Answer 2

`d/dx f(x)=10x^4+12x^3-3x^2-18x-15`

Explanation:

The product rule states that for a function, `f` such that;

`f(x) = g(x)h(x)`
`d/dx f(x) = g'(x)h(x) + g(x)h'(x)`

The function `f` is given as `f(x) = (x^3-3x)(2x^2+3x+5)`, which we can split into the product of two functions `g` and `h`, where;

`g(x) = x^3 - 3x`
`h(x) = 2x^2+3x+5`

By applying the power rule, we see that;

`g'(x) = 3x^2 - 3`
`h'(x)=4x+3`

Plugging `g`, `g'`, `h`, and `h'` into our power rule function we get;

`d/dx f(x) = (3x^2 - 3)(2x^2+3x+5) + (x^3 - 3x)(4x+3)`

`d/dx f(x)=6x^4+9x^3+15x^2-6x^2-9x-15+4x^4+3x^3-12x^2-9x`

`d/dx f(x)=10x^4+12x^3-3x^2-18x-15`