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Proof of Power Rule for 2 variables

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    u^n (x,y)=nu^(n-1) (x,y) u' (x,y)

    2. Relevant equations



    3. The attempt at a solution
    can i set f (x,y)=u^n (x,y)
    lnf =lnu^n
    lnf=nlnU
    f'/f=n/U
    f'=fn/U -(U(^n) )(n/U)
    after that i dont know how to continue and is there a better way to prove it
     
  2. jcsd
  3. Mar 4, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What do you mean by "u'(x,y)"? The gradient? The differential? "u' " is not defined for a function of two variables. If you mean the gradient, which is what I would think of as "the" derivative for a function of several variables, then yes, [itex]\nabla u^n(x,y)= nu^{n-1}\nabla u[/itex]. That follows from the chain rule.

     
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