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Homework Help: Product Rule

  1. Oct 23, 2005 #1
    How do I carry out the product rule for

    xy'z'

    Is it possible to do the product rule with y'z' and after that multiply it by x?
     
  2. jcsd
  3. Oct 23, 2005 #2
    What variable is this respect to? Is it [tex]\frac{d}{dx},\frac{d}{dy}[/tex],etc.
     
  4. Oct 23, 2005 #3
    It's not respect to any variable. It's just three separate variable... for instance it can be xyz.... so i was thinking that i first do the product rule for yz which is y'z + z'y and then i use this result and multiply it by x and x'. I am not sure if I am allowed to do this.
     
  5. Oct 23, 2005 #4
    The product rule extends for three variables as follows. Let's say we have three functions that are in terms of x, we'll call them f(x),g(x), and h(x).

    [tex]\frac{d(f(x)g(x)(h(x))}{dx}=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)[/tex].
     
  6. Oct 23, 2005 #5
    Thank you, I really appreciate your help.
     
  7. Feb 10, 2008 #6
    I have a question regarding the product rule.

    Our modern physics textbook asks us to derive the following:



    [tex]d(\gamma mu)=m(1- \frac{u^2}{c^2})^{-3/2} du[/tex]


    Is it implied that this is with respect to u? I can see the chain rule here, but I'm not sure precisely how this differentiation is done.
     
  8. Feb 10, 2008 #7

    HallsofIvy

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    Science Advisor

    First, please do not post a new question in someone else's thread. That is very rude- start your own thread.

    Second, strictly speaking, the right hand side is a differential with respect to u while the left side is just the differential of [itex]\gamma mu[/itex] and is not "with respect to" anything. If you were to rewrite it as
    [tex]\frac{d(\gamma mu)}{du}= m\left(1-\frac{u^2}{c^2}\right)^{-3/2}[/tex]
    then the derivative on the left is with respect to u.

    I can't tell you how to derive it since you haven't said what it is to be derived from- which, hopefully, you will do in a separate thread.
     
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