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Prove Power Rule by Math Induction and Product Rule

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the Principle of Mathematical Induction and the Product Rule to prove the Power Rule when n is a positive integer.

    2. Relevant equations

    Dxxn = nxn-1

    Dx(fg) = fDxg + Dxfg

    3. The attempt at a solution

    In summary,

    Dxxn = nxn-1
    Dxxk = kxk-1
    Dxxk+1 = (k+1)x(k+1)-1
    Dx(xkx) = (k+1)xk
    xkDxx + Dxxkx = (k+1)xk
    xk + kxk-1x = (k+1)xk
    xk + kxk = (k+1)xk
    (k+1)xk = (k+1)xk

    Therefore, Dxxn = nxn-1 is valid for all positive integers n.

    EDIT: Oh, and much appreciation for any help!
     
  2. jcsd
  3. Jul 21, 2011 #2

    Redbelly98

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    Looks like you have pretty much proved it. You just need to show it's true for n=1.
     
  4. Jul 21, 2011 #3

    gb7nash

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    If you really want to have a hoot, try proving the power rule from the definition of the derivative.
     
  5. Jul 21, 2011 #4

    hunt_mat

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    I pretty much did that in my lecture notes that I uploaded here.
     
  6. Jul 21, 2011 #5

    Redbelly98

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    It looks like the instructions are pretty clearly saying to use induction and the product rule:

     
  7. Jul 22, 2011 #6
    Did anyone read the proof yet? If I was the professor, I might give the proof half credit. I'm not intending in any way shape or form to criticize, 5hassay. That's why students come here: to learn and improve.

    On the third line of the summary, where k+1 is introduced, the proof appears to assume the result of the induction. Although for many proof methods, one can simply substitute what we need, this substitution is not allowed, because we are only allowed to assume it, the induction hypothesis, is true for k<=n.

    What about arranging the proof around
    xk+1= xk*x and then applying product rule to the right hand side?


    Also, clarify how you're going about this. Specifically, when doing induction stating the induction hypothesis, improves clarity.
    For example,
    Basis step: prove the statement for n=1.
    ... (easily forgotten/skipped, but essential)

    Inductive step:
    Assume for induction
    Dxxk= k*xk-1

    xk+1= xk*x
    Dxxk+1= Dx(xk*x) Take deriv. both sides
    Then apply product rule to right hand side and see what happens.
     
  8. Jul 22, 2011 #7

    SammyS

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    Pretty much what nickalh said.
     
  9. Jul 22, 2011 #8
    Ahhhh, thanks Sammy, I just realized OP actually worked it out acceptably on the left hand side. You're right, ironically, deleting the right hand side and cleaning up a bit does the proof. Possibly add a concluding line, I prefer actual English instead of just notation.

    Ya'll have an excellent day/afternoon/night.
     
  10. Jul 22, 2011 #9
    Haha, I do imagine that would be a good amount of fun and a good looking product afterward!
     
    Last edited: Jul 22, 2011
  11. Jul 22, 2011 #10
    Ah, okay. Yes, I am very new to Mathematical Induction (by definition), so I thought I would just skip the various statements. I now know that their importance is not less so as to not be involved in a summary! Also, I now understand about the n=1 step. I did indeed not think about it at all! Thanks for your help!
     
  12. Jul 22, 2011 #11
    Okay, thanks!
     
  13. Jul 22, 2011 #12

    SammyS

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    nickalh, I agree. I prefer English, rather than only notation, too.
     
  14. Jul 22, 2011 #13

    Redbelly98

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    You're welcome. Though what SammyS wrote in Post #7 is really correct. What you did was fine in terms of figuring things out, but a formally correct proof should follow along the lines of Post #7. Like I said, you pretty much have it.
     
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