So, I was given the following problem as part of a homework assignment.
Suppose $f'(x) > 0$ in $(a,b)$. Prove that $f$ is strictly increasing in $(a,b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that
$$g'(f(x)) = \frac{1}{f'(x)}$$
I have proven that $f$ is strictly increasing in $(a,b)$, and I could prove that $g'(f(x)) = 1/f'(x)$ if I could prove that $g$ is differentiable. The problem is that I am having trouble with a proof of that. Any advice?
Also, as a reference, this is exercise 5.2 from Baby Rudin.
Best Answer
Recall that if $f$ is invertible, then it is bijective. So, there exists a unique $y$ such that $y = f(x)$ on the domain of $g$, which is seen to be $(f(a),f(b))$ since we have an increasing function. By definition of the derivative:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(z) - g(f(x))}{z-f(x)} = \lim_{z \rightarrow f(x)} \frac{g(z) - x}{z-f(x)}$$
Now, as we tend $z$ closer and closer to $f(x)$, eventually it will have to belong to $(f(a),f(b))$, which means we can find another $x_z$ such that $f(x_z) = z$. We now choose $z$ sufficiently close, and take advantage of this fact. We then have:
$$g'(y) = \lim_{z \rightarrow f(x)} \frac{g(f(x_z)) - x}{f(x_z)-f(x)} = \lim_{z \rightarrow f(x)} \frac{x_z - x}{f(x_z)-f(x)}$$
We see that this final limit tends to $\frac{1}{f'(x)}$.