Question

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

Answer 1

See below.

Explanation:

The product rule: `d/dxu(x)v(x)=v(x)d/dxu(x)+u(x)d/dxv(x)`

Let `u(x)=(x+3)^2` and `v(x)=(x-1)`

Therefore `f'(x)` (just different notation to `d/dxf(x)`) is equal to:
`(x-1)d/dx(x+3)^2+(x+3)^2d/dx(x-1)`

Let's firstly simplify those derivatives:
`d/dx(x+3)^2` needs a use of the chain rule: `2(x+3)`
`d/dx(x-1)` is just `1`

Therefore, finishing off:
`f'(x)=2(x-1)(x+3)+(x+3)^2`
You could then factorise by (x+3):
`f'(x)=(x+3)[2(x-1)+(x+3)]`
Finally:
`f'(x)=(x+3)(3x+1)`