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Homework Help: Proof for exponential derivatives

  1. Dec 11, 2009 #1
    [tex]f(x) = 2^x \left \left[/tex]
    [tex]f(kx) = 2^(kx) \left \left[/tex]
    [tex]b = 2^k \left \left[/tex]
    [tex]b^x = 2^(kx) \left \left[/tex]
    [tex]b^x = f(kx)[/tex]
    [tex]\frac{d}{dx}(b^x) = \frac{d}{dx}(f(kx)) = \frac{d}{dx}(2^(kx)) (1)[/tex]
    [tex]\frac{d}{dx}(f(kx)) = k.f'(kx) (2)[/tex]
    I can't see how step (1) gets to step (2).
    Because I thought:
    [tex]\frac{d}{dx}(f(kx)) = k.\frac{d}{dx}(f(x))[/tex]
     
  2. jcsd
  3. Dec 11, 2009 #2
    I think it's:
    [tex]\frac{d}{d(kx)}(f(kx))*\frac{d}{dx}(kx) = k*f'(kx)[/tex]
     
  4. Dec 11, 2009 #3

    Mark44

    Staff: Mentor

    d/dx(f(u)) = d/du(f(u))*du/dx
     
  5. Dec 11, 2009 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    But isn't [tex] 2=e^{\ln 2} [/tex] ? Or am i missing something ?
     
  6. Dec 11, 2009 #5

    Mark44

    Staff: Mentor

    Yes, 2 = eln 2

    d/dx(f(kx)) = d/dx[(eln 2)kx] = ekxln 2 * k*ln2 = 2kx *k*ln2. The OP's formatting and organization made it a bit difficult to follow.
     
    Last edited: Dec 11, 2009
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