Hey, in my Schaum' Outline Calculus it says D(adsbygoogle = window.adsbygoogle || []).push({}); _{x}(e^{x}) = e^{x}

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

therefor 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}is

differentiable 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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# A more rigorous argument

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