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Differentiation math problem

  1. May 18, 2005 #1
    Quick question.

    To differentiate [tex]y = 20x (2x - 1)^6[/tex] I need to use the product rule:

    [tex] y = uv \ \Rightarrow \ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}[/tex]

    So that u = 20x and v = (2x-1)6

    However, do I differentiate v so that it equals 6(2x-1)5 or do I use:

    [tex]y = [f(x)]^n \ \Rightarrow \ \frac{dy}{dx} = n[f(x)]^{n-1} f'(x)[/tex]

    to give 12(2x-1)5???

    Cheers.

    The Bob (2004 ©)
     
    Last edited: May 18, 2005
  2. jcsd
  3. May 18, 2005 #2

    arildno

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    Why should it equal 6(2x-1)^5????
    You are to differentiate v with respect to "x", not with respect to "2x-1".
     
  4. May 18, 2005 #3
    Therefore it must be 12(2x-1)5???

    The Bob (2004 ©)
     
  5. May 18, 2005 #4

    arildno

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    Precisely, by the chain rule!
     
  6. May 18, 2005 #5
    Cheers.

    Sorry it seemed so trivial but I only started teaching myself tonight and the book I have sometimes misses some of the stages out e.g. I actually have v = (x-1)6 but this will come to 6(x-1)5 as the x-1 will become 1.

    Was simply checking.

    Appreciate the help.

    Cheers again. :biggrin:

    The Bob (2004 ©)
     
  7. May 18, 2005 #6

    dextercioby

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    It would be much easier to put Maple/other software to expand that bynomial make the multiplications and the then differentiate each term of the resulting sum.

    My guess...

    Daniel.
     
  8. May 18, 2005 #7

    arildno

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    Are you trying to bait me?
     
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