Hey, in my Schaum' Outline Calculus it says D_{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.
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.