# Can't decide if my proof is right or not ?inverse functions

1. Jul 24, 2013

### Andrax

so before i start this is extracted from spivak's calculus proof
now here is the problem the teacher who taught us limits and dervatives for the first time told us not to use f(x)-f(a) instead of f(a+h)-f(a)
so spivak wanted us to prove that if f a one-one function and is continious on an interval and differentiable on f$^{-1}$(b) then f'(b) exists and (f^-1)'(b)=$\frac{1}{f'(f^-1(b))}$(f^-1 = inverse of f )
i did the opposite of what spivak did , spivak started from the definition of f'(b) while i started wtih the definition of f$^{-1}$(b), so is my proof right?
let f(a) = b then f^-1(b)=a so
f'(f$^{-1}$(b) = f'(a) = $\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$=$\lim_{x \to f^-1(b)}\frac{f(x)-b}{x-f^-1(b)}$
now when x approches a=f^-1(b) in the domain of f , f(x)approaches f(a)=b in the domain on f-1 and we can set f(x)=x later on
so =$\lim_{f(x) \to b}\frac{f(x)-b}{x-f^-1(b)}$
replacing f(x) by the new x in the domain of f^-1 and b by f^-1(b) and x by f^-1(x)= a(I hope i'm right in this phase, please correct me if i'm wrong)
=$\lim_{x \to b}\frac{x-b}{f^-1(x)-f^-1(b)}$ = $\frac{1}{(f^-1)'(b)}$
i've always been confused when changing the x to f(x) : do we also change the function or it remains intact?Limx->af(x) <=>Limf(x)->f(a) [f(f(x))) or f(x)? ] if i made a fault it would be this, thanks for reading.

2. Jul 24, 2013

### verty

What is b? And you haven't named the interval. I think you need to be more precise.

3. Jul 24, 2013

### Andrax

It dosen't really matter with this proof but okay let b belongs to [n, m] where f is defined...

4. Jul 24, 2013

### micromass

I think your proof is intuitively right. However, an expression such as

$$\lim_{f(x)\rightarrow a}{g(x)}$$

is not formally defined. At least, I don't think Spivak defines it or works with it. So to get a formal proof, you either have to define precisely what the above means, or you need to write your proof to avoid the above notation.

5. Jul 24, 2013

### Andrax

Well yeah I'll try that I thought it's obvious since the pair x, f(x) is in f so f(x), x in f-1