Inverse Function Derivation proof

[mateomy]

I have a "pop" quiz tomorrow in my calc course and my professor is stating that we have to be able to restate the proof of inverse function derivatives. I am looking it over in my book and it's pretty straight forward except for one part that I can't figure out...
This is listed step by step

<br /> \ (f^-1)&#039;(a)=\lim_{x \to a}\frac{f^-1(x) - f^-1 (a)}{x-a}<br />

<br /> \lim_{y \to b}\frac{y-b}{f(y)-f(b)}<br />

<br /> \lim_{y \to b}\frac{1}{\frac{f(y)-f(b)}{y-b}}=\frac{1}{\lim_{y \to b}\frac{f(y)-f(b)}{y-b}}<br />

<br /> = \frac{1}{f&#039;(b)} = \frac{1}{f&#039;(f^-1(a))}<br />

I get lost on the second to last line where it switches from the f(y)-f(b)/y-b to just the reciprocal of f'(b). The former is simply just the definition of the derivative of the function b, right?

 

[LeonhardEuler]

<br /> \lim_{y \to b}\frac{1}{\frac{f(y)-f(b)}{y-b}}=\frac{1}{\lim_{y \to b}}\frac{f(y)-f(b)}{y-b}<br />

This should be
\lim_{y \to b}\frac{1}{\frac{f(y)-f(b)}{y-b}}=\frac{1}{\lim_{y \to b}\frac{f(y)-f(b)}{y-b}}
right?
edit: Ok, you fixed it. Yes the limit on the bottom is the definition of the derivative of 'f' at b. What is your question exactly?

 

[mateomy]

Yeah, I just fixed it and saw your post; but yeah, it looks accurate to the book now.

 

[LeonhardEuler]

I get lost on the second to last line where it switches from the f(y)-f(b)/y-b to just the reciprocal of f'(b). The former is simply just the definition of the derivative of the function b, right?

Ok, what do you mean by "the function b"; Do you mean f'(b)? That is what it means and that's what they substitute, so it works out.

 

[mateomy]

Yes, I meant f'(b). After typing up the latex (strangely enough) it sort of pushed its way through in my mind. But the confirmation definitely helps, thank you for your time.