Expert: Paul Klarreich - 3/14/2006

Question
1.  Deborah 2. Calculus 3. I've already tried the problem but dont understand how to arrrive at the answer.  Please explain the steps for the first derivative to: (2t^2-1)^2 (3t^2).  I get as far as:
2(2t^2-1)(4t)(3t^2) - (2t^2-1)^2 (6t) then what are the next steps?
The answer is: 6t(12t^4-8t^2 +1) Please help.


Answer
Hi, Deborah,
You wrote:
Subject:  Derivative
Question:  1. Deborah 2. Calculus 3. I've already tried the problem but dont understand how to arrrive at the answer. Please explain the steps for the first derivative to: (2t^2-1)^2 (3t^2). I get as far as:
2(2t^2-1)(4t)(3t^2) - (2t^2-1)^2 (6t) then what are the next steps?
The answer is: 6t(12t^4-8t^2 +1) Please help.
------------------------------
You have:
y = (2t^2 - 1)^2 (3t^2)

Use the Product rule:

y' = 2(2t^2 - 1)(4t)(3t^2) + (2t^2 - 1)^2(6t)

OK. You got this.  Next steps are simple algebra.  (Yeah, right! Simple, he says.)

Multiply out each term.  Maybe you use a separate piece of paper for each, then put things together.  Afterwards.

2(2t^2 - 1)(4t)(3t^2) = 24t^3(2t^2 - 1) =
48t^5 - 24t^3  [FIRST PART]

(2t^2 - 1)^2(6t) = (4t^4 - 4t^2 + 1)(6t) =  
24t^5 - 24t^3 + 6t  [SECOND PART]

Now: [USE A FIXED-SIZE FONT TO VIEW THIS.]

      FIRST            SECOND
    -------------   ------------------
y' = 48t^5 - 24t^3 + 24t^5 - 24t^3 + 6t

Combine like terms:

y' = 72t^5 - 48t^3 + 6t

Remove a common factor of 6t:

y' = 6t(12^4 - 8t^2 + 1)

That's it.