How can I find the second derivative of f(x) = 12x(x-1)^3?

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Homework Help Overview

The discussion revolves around finding the second derivative of the function f(x) = 12x(x-1)^3. Participants are exploring the process of differentiation, particularly focusing on the application of product and chain rules in calculus.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the second derivative after successfully calculating the first derivative. They express confusion regarding their calculations and the discrepancy between their result and the expected answer. Other participants question specific steps in the differentiation process and offer clarifications on applying the product and chain rules correctly.

Discussion Status

Participants are actively engaging in clarifying the steps involved in finding the second derivative. Some guidance has been provided regarding the correct application of differentiation rules, and there is an ongoing exploration of how to factor and simplify the results. Multiple interpretations of the steps are being discussed without reaching a consensus.

Contextual Notes

There is a noted confusion regarding the handling of the squared term (x-1)^2 and how it affects the derivative calculations. Participants are also questioning how to factor expressions correctly to arrive at the expected form of the second derivative.

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
 
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Hey mdklimer,

I see where the error is in your calculations. After you do the product and the chain rule you somehow got 96(x-1) and with the chain/product rule, there is no way you could have come to that. (x-1) is squared, not mulitplied by 2.

After that, you should be able to arrive at the right answer :)
 
I got 96(x-1) by bringing down the two in (x-1)^2, and multiplied it by the 48 next to it. Okay, so maybe that is wrong, but I still don't see how the end result is 72(2x-1)(x-1).
Then what do you do with the (x-1)^2?

Also, they somehow get (2x-1) from my (4x-1), how is that possible? Do you factor out a 2 from (4x-1)?
 
You don't bring down the 2 in (x-1) because you only need to get the derivative of 48x-12, t(x-1)^2 stays how it is because the product rule says that the derivative of two functions is the first times the derivative of the second, plus the derivative of the first times the second function. so you'll get 2(x-1)(48x-12) + (48)(x-1)^2. What you can do after that is factor out the 12 in 48x-12 and get (24)(x-1)(4x-1)+(48)(x-1)^2, because 2 times 12 gives you 24.

After this you can do many other different methods, like multiplying everything and then simplifying, but that is very tedious, atleast for me. What I did is factored out 24(x-1) from the whole equation to get:
24(x-1)((4x-1)+2(x-1))
Once you start simplifying you will see that you'll end up with 72(2x-1)(x-1)
 
Wow, thanks so much. That explains a lot about what I was doing wrong. I now understand that the factoring and simplifying was my problem. Nice work Priscila, thanks again.
 
You're welcome, I'm glad I was able to help.
 

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