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Extended product rule for derivatives

  1. Nov 22, 2008 #1
    Can someone please explain it to me? My handwriting wasn't at its best when I was taking notes in class and now I can't read it. The teacher showed an example that I jotted down but what's the general rule?
     
  2. jcsd
  3. Nov 22, 2008 #2
    Maybe you should consult your textbook or take notes more carefully.
    The usual product rule for differentiation reads
    [tex]
    (uv)'=uv'+u'v
    [/tex]
    where u,v are functions.

    What do you mean by extended product rule?
     
  4. Nov 22, 2008 #3
    I meant if you have 3 or more terms, like y = sinxcosxlnx and you want the derivative.
     
  5. Nov 22, 2008 #4

    cristo

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    For three terms: (uvw)'=(uv)'w+(uv)w'=u'vw+uv'w+uvw'.

    As you can see, this can be extended to 4 terms and beyond quite readily.
     
  6. Nov 22, 2008 #5

    nicksauce

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    It's easy to derive yourself, using the regular problem rule. If you have y = f(x)g(x)h(x), then use the formula Pete Callahan posted, using u=f(x), v=(g)h(x).
    So then y' = uv' + vu'
    Then use the product rule again to find v'
     
  7. Nov 22, 2008 #6
    Thanks everyone. This makes more sense now.
     
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