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Differentiation & composite functions (2)

  1. Oct 23, 2008 #1
    Suppose we are given that
    d
    --F(y) = f(y)
    dy

    Then is it true that
    dF(h(u))
    -------- = f(h(u)) dh(u)/du ?
    du
    Why or why not? In particular, I don't understand how to get the red part...

    Thanks for explaining!
     
    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 23, 2008 #2

    CompuChip

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    It is basically just application of the chain rule.
    Calling y = h(u),

    [tex]\frac{dF(y)}{du} = \frac{dF(y)}{dy} \times \frac{dy}{du}[/tex]
    so if you call dF(y)/dy = f(y) then you have your identity (just replace y with its definition h(u) again).
     
  4. Oct 23, 2008 #3

    Ben Niehoff

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    Try this: http://en.wikipedia.org/wiki/Chain_rule

    There is a proof about halfway down the page.

    If that proof is confusing, then I should think your calculus textbook ought to provide a proof also, which may be easier to follow.

    Or you can try to prove it yourself using the limit definition of the derivative.
     
  5. Oct 23, 2008 #4
    OK, thanks!

    But can we replace y by h(u) and vice versa that freely?

    Sorry my calculus is a bit rusty now...
     
  6. Oct 24, 2008 #5

    HallsofIvy

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    You can replace anything by anything that is equal to it!
     
  7. Oct 24, 2008 #6

    CompuChip

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    I defined y to be h(u). So, as HallsOfIvy points out, anywhere you see one you can replace it by the other.
     
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