A more rigorous argument

MaxManus

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(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.

HallsofIvy

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.

MaxManus

Thanks