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

1 Answer
May 29, 2016

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)