Proving that the derivative of the inverse exists Let $f$ be differentiable and invertible around $a, f(a) = b, f(a + h) = b + k$.
For any function $G$, each of the following is true:
1) $\lim_{h\to 0}G(h)=L\iff \lim_{k\to 0}G(h)=L$
2) $\lim_{h\to 0}\frac{G(h)}{h}=0\iff \lim_{k\to 0}\frac{G(h)}{k}=0$  
How can I prove that $(f^{−1})′(b)$ exists?
It's question 4(d) on page 2 of this document.  
EDIT:
The problem sets at the link considers the situation $ f(a) = b, f(a + h) = b + k$, and tries to determine under what conditions the claim $k \rightarrow 0$ is equivalent to $h \rightarrow 0$.
The original question 4 assumes $f$ is continuously differentiable and $f'(a)\neq 0$,then in some neighborhood U of a, $mh \le k \le Mh$ ,where either both $m,M >0$ or both $m,M<0$, by the Mean Value Theorem. Thus in U, there is a one-to-one correspondence between values of h and values of k.  
In sub-questions from a) to c), the following is asked to be proved using $\epsilon-\delta$ and assuming $mh \le k \le Mh$ $(m,M > 0)$ , for any function G:  
a) $\lim_{h\to 0}G(h)=L\iff \lim_{k\to 0}G(h)=L$
b) $\lim_{h\to 0}\frac{G(h)}{h}=0\iff lim_{h\to 0}\frac{G(h)}{k}=0$
c) $\lim_{h\to 0}\frac{G(h)}{h}=0\iff \lim_{k\to 0}\frac{G(h)}{k}=0$  
Sub-question d) asks what the relevance of the earlier parts of the question to the proof that if f is invertible around a, and $f'(a)\neq 0$, then  $(f^{−1})′(b)$ exists is.  
Is the part of this post before EDIT what sub-question d) is asking ? How can I solve sub-question d) ? Any idea is appreciated.
 A: The derivative of the inverse exists if and only if $f'(a) \neq 0$ and then we have $(f^{-1})'(b)=1/f'(a)$.

Note that the variable $k$ defined by $$f(a+h) =f(a) +k\tag{1}$$ tends to $0$ as $h\to 0$ and at the same time $k\neq 0$ as $h\to 0$ because $f$ is continuous and invertible and hence one-one around $a$. 
Now let $g$ be the inverse of $f$ and $g(b) =a$. Since $f$ is continuous around $a$,  it follows that $g$ is continuous around $b$. Now we can see that if we define $$g(b+k) =g(b) +h\tag{2}$$ then this is equivalent to the earlier equation $(1)$ (just apply $f$ to both sides of $(2)$). Since $k/h\to f'(a) $ as $h\to 0$ and $f'(a) \neq 0$ it follows that $h/k\to 1/f'(a)$ as $h\to 0$. From equation $(2)$ and continuity of $g$ it follows that as $k\to 0$ we must have $h\to 0$ and by $(1)$ we have $h\neq 0$. Thus we can say that $h/k\to 1/f'(a)$ as $k\to 0$. Thus we can see that $g'(b) =1/f'(a)$.

I don't think you need complicated steps like a, b, c and $\epsilon, \delta$ is an overkill for this problem. The important thing to note here is that as $h\to 0$ we have $k\to 0,k\neq 0$ and as $k\to 0$ we have $h\to 0,h\neq 0$. Both these assertions follow from the fact that $f$ is continuous and invertible around $a$.

Further note the following result which follows directly from the definition of limit

Let the variables $h, k$ be connected by a functional relation such that as $h\to 0$ we have $k\to 0,k\neq 0$ and as $k\to 0$ we have $h\to 0,h\neq 0$. Then the limit operation $\lim_{h\to 0}$ can be replaced by limit operation $\lim_{k\to 0}$ and vice versa.

See this answer to a related question. 
