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Proving the power rule by induction

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data

    If a is a natural number, prove by induction that

    y = [g(x)]^a => y' = a[g(x)]^(a-1) * g'(x)

    2. Relevant equations


    Let a = 2

    y' = (2)[g(x)]^(2-1) g(x)
    = 2g(x)g'(x)

    Let a = 3

    y' = (3)[g(x)]^(3-1) g(x)
    = 3g(x)^2 * g'(x)

    Let k be any natural number

    a(k) = y' = ak[g(x)]^(ak-1) * g'(x)

    3. The attempt at a solution

    What I did in the above equation was to substitute 2 and 3 (both natural numbers) as a, in order to prove that every natural number k is applicable.

    I'm not all too familiar with induction, but am I on the right track? Or am I completely off?

    How do I prove the power rule through induction?
     
  2. jcsd
  3. Apr 27, 2009 #2

    dx

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    Gold Member

    This is how induction works:

    1. First, you prove that it is true for a = 1.

    2. Then, you assume that it is true for a = k, and then show that this implies that it is true for a = k + 1.

    This is enough to show that it is true for all natural numbers.
     
  4. May 3, 2009 #3

    HallsofIvy

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    Science Advisor

    Why a= 2 and 3? You haven't shown the statement to be true for 1 and that's the critical starting point.

    This is what you were supposed to prove. You are simply asserting it.


    What you want to prove is:
    If for some k, [itex](ag(x)^k)'= ag(x)^{k-1}g'(x)[/itex] then [itex](ag(x)^{k+1})'= ag(x)^k g'(x)[/itex].

    That is, if the statement is true for k, it is true for k+1. Then, if you have also proved it true for k= 1, it must true for k+1= 2. Then, since it is true for k= 2, it is true for k+1= 3, etc.

    Try writing [itex](g(x))^{k+1}[/itex] as [itex]g(x)(g(x))^k[/itex] and use the product rule.
     
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