mdklimer
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I have this problem for homework dealing with second derivatives and graphs. I have no problem finding derivatives usually, but this one is giving me trouble. I cannot figure out how to get the second derivative. I have an idea of what to do, but need some extra guidance.
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
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