A more rigorous argument

  1. Hey, in my Schaum' Outline Calculus it says Dx(ex) = ex
    Let y = ex. Then ln(y) = x. By implicit differentiation, [tex]\frac{1}{y}[/tex]y' = 1
    therefor y' = y = ex


    For a more rigorous argument, let f(x) = ln(x) and f-1(y) = ey.
    Note that f'(x) = [tex]\frac{1}{x}[/tex]. By Theorem 10.2(b).

    (f-1)'(y) = [tex]\frac{1}{f'(f**-1(y))}[/tex],

    That is Dy = [tex]\frac{1}{1/e**y}[/tex] = ey

    10.2(b): Let f be one-to-one and continuous on the interval (a,b) Then:
    If f'x(x0 is differentiable and f'(x0) != 0, then f-1 is
    differentiable at y0 = f(x0) and (f-1)'(y0)
    = [tex]\frac{1}{f'(x0}[/tex]

    Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous.
     
    Last edited: Jan 9, 2009
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,673
    Staff Emeritus
    Science Advisor

    The first argument uses the fact (your "10.2(b)") that if y= f(x) and x= f-1(y) then
    [tex]\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex]
    The second argument proves that is true for this particular function, as part of the proof.
     
  4. Thanks
     
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