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Chain rule problem: proper method?

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the chain rule, the derivative formula Dxsinu=cosuDxu, together with the identities
    cosx=sin([itex]\pi[/itex]/2 -x) and sinx=cos([itex]\pi[/itex]/2 -x)
    to obtain the fomula for Dxcosx.


    2. Relevant equations

    Chain rule: dy/dx=dy/du[itex]\cdot[/itex]du/dx


    3. The attempt at a solution

    For my second attempt (the website made me log in again when I clicked on "Preview Post" last time - that seems to happen a lot):
    Dxcosx=Dxsin([itex]\pi[/itex]/2 -x)
    =cos([itex]\pi[/itex]/2 -x)(-1)
    =-sinx

    This is the result that I expected, but the formula in the text book says Dxcosu=-sinuDxu. I realize that this formula contains "u"s and the required formula does not. I also realize that Dxx=1, which does account for the discrepancy between the answer that I got and the formula in the book, but I can't help wondering if it was by sheer happenstance that it turned out to work. I see no progression that could lead me to obtain the result that Dxcosx=-sinxDxx.

    I'm not sure if I'm employing the chain rule properly. As you can see, I only used one variable in my proof, but the question uses both "u" and "x." Is there some form of notation that is considered correct for this kind of proof? (If so, I'd love to know why. There seem to be too many ways of expressing the same idea: dy/dx, d/dx, Dx, f'(x), others?) Thanks for your input.
     
  2. jcsd
  3. Jul 4, 2011 #2

    lanedance

    User Avatar
    Homework Helper

    your derivative is correct, the only difference is the book leave it in a general form, for a general function u(x)

    notation should go with the context of the problem, or how you learn it in class I guess, but all those you mention can mean the same thing. \

    Generally if there's any chance of confusion I try and be as explicit as possible
    [tex]\frac{dy(x)}{dx}[/tex]
     
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