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Mathematical Induction: Power Rule for Differentiation

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that
    [tex]\frac{d}{dz}z^n = nz^{n-1}\;\;\; \forall n\in\mathbb{N}[/tex]
    using the Product Rule for differentiation and mathematical induction.

    2. Relevant equations
    [tex]\frac{d}{dz} f(z) = \lim_{\Delta z\to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z}[/tex]
    [tex]\frac{d}{dz}[f(z)g(z)] = f\,'(z)g(z) + g'(z)f(z)[/tex]


    3. The attempt at a solution
    Let n = 1:
    [tex]\frac{d}{dz} z = \lim_{\Delta z \to 0} \frac{z + \Delta z - z}{\Delta z} = \lim_{\Delta z \to 0} \frac{\Delta z}{\Delta z} = \lim_{\Delta z \to 0} 1 = 1[/tex]

    Assume true for n = k:
    [tex]\frac{d}{dz} z^k = kz^{k-1}[/tex]

    Let n = k + 1:
    [tex]\frac{d}{dz} z^{k+1} = \frac{d}{dz}[z^k\cdot z] = \frac{d}{dz} (z^k) \cdot z + \frac{d}{dz} (z) \cdot z^k = kz^{k-1}\cdot z + z^k = kz^k + z^k = (k+1)z^k = (k+1)z^{(k+1)-1}[/tex]

    Like my previous topic, I'm pretty sure I have the proof correct, but I need to make sure that it is written out correctly. I don't want to be teaching myself the wrong way to lay out proofs.
     
  2. jcsd
  3. Jul 8, 2011 #2

    LCKurtz

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    Looks fine to me.
     
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