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

1 Answer
May 29, 2016

Answer:

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)#