# A more rigorous argument

1. Jan 9, 2009

### MaxManus

Hey, in my Schaum' Outline Calculus it says Dx(ex) = ex
Let y = ex. Then ln(y) = x. By implicit differentiation, $$\frac{1}{y}$$y' = 1
therefor y' = y = ex

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

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

That is Dy = $$\frac{1}{1/e**y}$$ = ey

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.

Last edited: Jan 9, 2009
2. Jan 9, 2009

### 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.

3. Jan 10, 2009

Thanks