What makes the second argument in Schaum's Outline Calculus more rigorous?

[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, $$\frac{1}{y}$$y' = 1
therefor y' = y = e^xFor a more rigorous argument, let f(x) = ln(x) and f^-1(y) = e^y.
Note that f'(x) = $$\frac{1}{x}$$. By Theorem 10.2(b).

(f^-1)'(y) = $$\frac{1}{f'(f**-1(y))}$$,

That is D_y = $$\frac{1}{1/e**y}$$ = 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)
= $$\frac{1}{f'(x0}$$

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
$$\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}$$
The second argument proves that is true for this particular function, as part of the proof.

 

[MaxManus]

Thanks