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## Main Question or Discussion Point

Hey, in my Schaum' Outline Calculus it says D

Let y = e

therefor y' = y = e

For a more rigorous argument, let f(x) = ln(x) and f

Note that f'(x) = [tex]\frac{1}{x}[/tex]. By Theorem 10.2(b).

(f

That is D

10.2(b): Let f be one-to-one and continuous on the interval (a,b) Then:

If f'x(x

differentiable at y

= [tex]\frac{1}{f'(x0}[/tex]

Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous.

_{x}(e^{x}) = e^{x}Let y = e

^{x}. Then ln(y) = x. By implicit differentiation, [tex]\frac{1}{y}[/tex]y' = 1therefor y' = y = e

^{x}For a more rigorous argument, let f(x) = ln(x) and f

^{-1}(y) = e^{y}.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 D

_{y}= [tex]\frac{1}{1/e**y}[/tex] = e^{y}10.2(b): Let f be one-to-one and continuous on the interval (a,b) Then:

If f'x(x

_{0}is differentiable and f'(x_{0}) != 0, then f^{-1}isdifferentiable at y

_{0}= f(x_{0}) and (f^{-1})'(y_{0})= [tex]\frac{1}{f'(x0}[/tex]

Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous.

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