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Math Help - Chain rule problem

  1. #1
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    Chain rule problem

    Hey helpful member of this forum, I am currently in a calculus and vectors self learning course, and I need help with a problem. ***As I wrote this thread I was doing all the math again. I found that my problem is my finding of the derivative of  y = u(u^2 + 3)^3 . I kept the rest of my solution up just incase this is also a problem. My main concern, however, is my bold statement showing where the main problem (I think) is.***

    Question: Use the chain rule to find dy/dx at the indicated value of x.
     y = u(u^2 + 3)^3, u = (x + 3)^2, x = -2

    My solution:
    Basically how the text explains the chain rule is: "If f and g are functions having derivatives, then the composite function  h(x) = f(g(x)) has a derivative given by  h'(x) = f'(g(x))g'(x) "

    So with this information this is how my solution looks:
    First I expanded  u = (x+3)^2 :
     u = x^2 + 6x + 9

    Which makes it easier for me to find the derivative (just me likely):
     du/dx = (1)(2)x^{2-1} + (1)(6)x^{1-1}
     = 2x + 6

    I believe this is where the problem lies!!! So help in finding the derivative here would be awesome!!!
    Then I found the derivative of  y = u(u^2 + 3)^3 :
     dy/du = (u)(3)(u^2 + 3)^{3-1}((1)(2)u^{2-1})
     dy/du = 3u(u^2 + 3)^2(2u)
     dy/du = 6u^2(u^2 + 3)^2 *
    *According to wolfram alpha, this derivative is incorrect. I do not understand why. The solution on wolfram alpha when I input  y = u(u^2 + 3)^2 is  (3 + u^2)^2 (3 + 7u^2) . I do not understand how they come to this answer.* Link: http://www.wolframalpha.com/input/?i=y+%3D+u%28u^2+%2B+3%29^3

    Assuming the above is correct, subbing  u = (x + 3)^2 into  y = u(u^2 + 3)^3 should *assuming the answer in the book is correct* come to 320 when I sub in -2 for x. So:
     (6((x)^2 + 6(x) + 9)^2)((((x)^2 + 6(x) + 9)^2) +3)^2
    Then I sub in x = -2:
    (6((-2)^2 + 6(-2) + 9)^2)((((-2)^2 + 6(-2) + 9)^2) +3)^2
     = (6((1)^2))(4)^2
     = (6)(16)
     = 96 Then multiply by 2x + 6:
     (96)(2)
     = 192


    When I use the derivative shown on wolfram alpha I do get 320. *I tried this before but did not get correct answer, just tried again now and did!!!* How do I find the derivative shown on wolfram alpha?
     (3 + (x^2 + 6x + 9)^2)^2 * (3 + 7(x^2 + 6x + 9)^2)
    Sub in x = -2:
     ( 3 + ((-2)^2 + 6(-2) + 9)^2)^2 * (3 + 7((-2)^2 + 6(-2) + 9)^2)
     = ((3 + 1)^2)(3 + 7(1))
     = (16)(10)
     = (160)(2) *multiply by (2x + 6) with x = -2 subbed in*
     = 320


    - Thanks for looking at my thread!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    another way of thinking of the chain rule is

    If y = y (u) and u = u(x)

    Then dy/dx = (dy/du)(du/dx)

    For y= u (u^2 + 3) ^3 where u = (x+3)^2

    dy/du = (u^2 + 3) ^3 + 6u^2 (u^2 + 3) ^2 (Product rule)

    dy/du = [(u^2+3)^2] [7u^2 +3]

    du/dx = 2(x+3)

    dy/dx = 2[(u^2+3)^2] [7u^2 +3](x+3)

    Normally we would now convert u to (x+3)^2 but we only need the derivative when x= -2 . Note if x= -2 then u = 1

    substitute into the above eqn.
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