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Proof by Induction of the Power Rule of Differentiation

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Okay, the concept here is to use induction to prove that for n, (f1 x f2 x ..... x fn-1 x fn)' = (f'1 x f2 x ... x fn) + (f1 x f'2 x ... x fn) + .... + (f1 x f2 x ... x f'n).

    2. Relevant equations/ 3. The attempt at a solution

    I solved the initial step, which was quite easy. I started to set up the inductive step, by stating that:

    (f1 x f2 ..... x fn x fn+1)' = (f'1 x f2 x ... x fn+1) + (f1 x f'2 x ... x fn+1) + .... + (f1 x f2 x ... x f'n+1).

    And I do understand how induction works - I know I am supposed to plug in what I have for the "n" equation into part of my "n+1" equation and find equality. I just don't know HOW I'm supposed to do that for some reason. I think it's the derivatives throwing me off because in the last few problems that I did, it was fine. Please help me simplify?
     
  2. jcsd
  3. Nov 20, 2012 #2

    Dick

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    (f1 x f2 ..... x fn x fn+1)=(f1 x f2 ..... x fn) x fn+1. Use the usual two factor product rule on that.
     
  4. Nov 20, 2012 #3
    Okay, that makes it simpler. I didn't know if I was "allowed" to assume that, or if I had to go from the ground up. I'll double-check with my professor but for now I'll work it out that way. Thanks!
     
  5. Nov 20, 2012 #4

    Dick

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    I didn't "assume" anything. (fg)'=f'g+fg' is the base case for your induction. It's the usual product rule. Sure, you can assume that. And (f1 x f2 ..... x fn x fn+1)=(f1 x f2 ..... x fn) x fn+1 is just using the associative rule for multiplication. I think you are pretty safe in using that!
     
  6. Nov 21, 2012 #5

    HallsofIvy

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    You titled this "power rule" but talk about the "product rule". Which are you trying to prove?
     
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