How was the algebraic manipulation in this proof done?

Question

There is a proof in my math book about how:

$g'(f(x))=\frac{1}{f'(x)}$ where $g(x)=f^{-1}$

The author did this by using the definition of derivative in the following way:

Let $k(h)=g(f(x)+h)-g(f(x))$

This means that:

$g'(f(x))=\lim_{h\to 0} \frac{g(f(x) + h) - g(f(x))}{h} = \lim_{h\to 0} \frac{k(h)}{f(x+k(h))-f(x)} = \lim_{k\to 0} \frac{k}{f(x+k)-f(x)} =\frac{1}{f'(x)}$

It might be really simple, but I dont see how $\lim_{h\to 0} \frac{g(f(x) + h) - g(f(x))}{h}$ was rewritten to $\lim_{h\to 0} \frac{k(h)}{f(x+k(h))-f(x)}$

Answer 1

$$h=f(x+k(h))-f(x)\impliedby$$ $$f(x)+h=f(x+k(h))\impliedby$$ $$g(f(x)+h)=x+k(h)\impliedby$$ $$g(f(x)+h)-x=k(h).$$

Answer 2

As a little bit of a beginner myself, I find it a little hard to understand the first answer, so I am adding this answer which has more explanation and can be understood by everyone.

Since g(x) is the inverse function of f(x), this means that f(g(x)) = g(f(x)) = x.

So k(h) = g(f(x)+h) - g(f(x)) = g(f(x)+h) - x

Adding x on both sides to get x + k(h) = g(f(x)+h)

Thus, denominator of RHS = f(x + k(h)) - f(x) = f(g(f(x) + h)) - f(x) = (f(x) + h) - f(x) = h(remember that f(g(x)) = x)

This is the denominator of LHS

And the numerator of RHS is defined to be k(h) only