- #1

- 3

- 0

Find f '(x) where f(x) = 12x(x-1)^3

I have no trouble getting the first derivative which is:

f '(x) = 12(x-1)^2 (4x-1)

Now the next step is to find f ''(x) where f '(x) = 12(x-1)^2 (4x-1)

The end answer from the book is f ''(x) = 72(2x-1)(x-1).

I don't get any where near of what they get. Here are my steps in trying to solve:

f '(x) = 12(x-1)^2 (4x-1)

= (x-1)^2 (48x-12) <==== Multiplied 12 with (4x-1) to make it into two factors.

= 2(x-1) (1) (48x-12) + (48)(x-1)^2 <==== Product/Chain Rule

= 2(x-1) (48x-12) + 96(x-1)

= 2(x-1) 12(4x-1) + 96(x-1)

= 24(x-1)(4x-1) + 96(x-1)

f ''(x) = 120(x-1)(4x-1)

Again, the end answer from the book is f ''(x) = 72(2x-1)(x-1)

I figure there is some way to factor my answer more, but I don't know how.

Any help would be appreciated.

Thanks