# A more rigorous argument

by MaxManus
Tags: argument, rigorous
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 P: 297 Hey, in my Schaum' Outline Calculus it says Dx(ex) = ex Let y = ex. Then ln(y) = x. By implicit differentiation, $$\frac{1}{y}$$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) = $$\frac{1}{x}$$. By Theorem 10.2(b). (f-1)'(y) = $$\frac{1}{f'(f**-1(y))}$$, That is Dy = $$\frac{1}{1/e**y}$$ = 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) = $$\frac{1}{f'(x0}$$ Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 The first argument uses the fact (your "10.2(b)") that if y= f(x) and x= f-1(y) then $$\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}$$ The second argument proves that is true for this particular function, as part of the proof.
 P: 297 Thanks

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