Inverse function differentiability proof My textbook states the proof as follows:
Let $f$ be a continuous one-to-one function defined on an interval, and suppose that $f$ is differentiable at $f^{-1}(b)$, with derivative $f'(f^{-1}(b)) \not = 0$. Then $f^{-1}$ is differentiable at $b$, and 
$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$.
Let $b = f(a)$. Then $\lim_{h \to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \lim_{h \to 0} \dfrac{f^{-1}(b + h) - a}{h} $
Now every number $b + h$ in the domain of $f^{-1}$ can be written in the form $b + h = f(a + k)$ for a unique $k$ (we should write $k(h)$, but we will stick with $k$ for simplicity).

I am confused with two parts: 


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*Why can every number $b + h$ in the domain of $f^{-1}$ be written in the form $b + h = f(a + k)$ for a unique $k$? What is the reasoning behind this?

*Why should we write $k(h)$?



I would greatly appreciate it if people could please take the time to clarify this.

Okay, let's look at these questions one by one:

  
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*Why can every number $b+h$ in the domain of $f^{−1}$ can be written in the form $b+h=f(a+k)$ for a unique $k$? What is the reasoning behind this?
  

Well, we know that $f^{-1}$ is defined on the range of $f$. We can conclude from the intermediate value theorem that f is monotonic. 
So, for any $y \in Dom(f^{-1})$ we have some $x$ such that $f(x)$ = y. Now fix $a$ , $b$ such that $f(a) = b$. 
So for $b+h \in Dom(f^{-1})$ we have a unique $x$ such that f(x) = b+h. This $x$ can now be uniquely written as $a + k$ for some $k$.


  
*Why should we write $k(h)$?
  

This one is easy. All it says is that whatever k you choose such that $$f(a+k) = b+h$$ depends on h. For different values of h, one has different values of k that satisfy the equation above. Remember, we look at h and k after having chosen fixed points a and b
