Using properties of inverse functions to prove a derivative exists. A quick differentiation question that I'm caught up on:
Let $n \in \mathbb{N}$ and let $A=[0,\infty)$ if $n$ is even, and $A=\mathbb{R}$ if $n$ is odd. Let $g: A\rightarrow\mathbb{R}$ be $g(y)=y^{1/n}.$
Prove: For $y \neq 0$, $g'(y)$ exists and $g'(y)=\frac{1}{n}y^{\frac{1}{n}-1}$.*
Thus far, I've tried to define the inverse function $f(y)=y^n$, then using the fact that if $g(f(y))=y$ then $g'(f(y))=\frac{1}{f'(x)}$.
Though I think I have to show that $g$ is continuous, and $f$ differentiable to be able to use this result.
If we show this, do we then get that
$$g'(y)= \lim_{f(x) \to f(y)} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} = \lim_{f(x) \rightarrow f(y)}\frac{x-y}{f(x)-f(y)}=\lim_{f(x) \rightarrow f(y)}\frac{x-y}{x^n-y^n}?$$
I'm not quite sure how to approach this question. Any help is appreciated.
 A: If we want to talk about inverse functions, then we need to restrict the codomain so as to make the function invertible: Instead of $g : A \to \Bbb R$, we need $g : A \to A$. Then by the definition of $y^{1/n}, g : A \to A$ is the inverse of $f : A \to A$.

Though I think I have to show that $g$ is continuous, and $f$ differentiable to be able to use this result.

There are many variants of the theorem. I don't know which one you are referring to. But you should look it up and make sure you know its requirements.
That $f$ is differentiable is surely something you've already seen in your course. If nothing else, it follows by induction from the Leibnitz rule $(uv)' = u'v + v'u$.
That $g$ is continuous follows from it being the inverse of a continuous function on an interval. Because $f$ is both continuous and invertible, it has to be strictly monotone. (Otherwise it would violate the intermediate value theorem - do you see why?) Therefore $g$ is also strictly monotone and can only have jump discontinuities. Such a discontinuity would mean that $f$ was not defined in the jumped region, but $f$ is defined everywhere in the interval.

If we show this, do we then get that
$$g'(y)=\lim_{f(x) \to f(y)} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} =\lim_{f(x) \to f(y)}\frac{x-y}{f(x)-f(y)}=\lim_{f(x) \to f(y)}\frac{x-y}{x^n-y^n}$$

No. The definition of $g'(y)$ is $g'(y) = \lim_{x \to y}\frac{g(x) - g(y)}{x-y}$. If you already knew that $g'(y)$ existed, then you could claim that $\lim_{x \to y} \frac{g(f(x))-g(f(y))}{f(x)-f(y)} = g'(y)$. But the existence of $\lim_{x \to y} \frac{g(f(x))-g(f(y))}{f(x)-f(y)}$ does not in general imply the existence of $\lim_{x \to y}\frac{g(x) - g(y)}{x-y}$, since $f$ might be sidestepping where $g$ misbehaves.
And note that I am still taking the limit as $x \to y$. You would have to define what the limit as "$f(x) \to f(y)$" even means. Because this particular $f$ is well-behaved, you could get away with doing so, but in general "$\lim_{f(x) \to f(y)}$" is not a real thing.
However, you do get
$$\lim_{x\to y} \frac{g(x) - g(y)}{x-y}=\lim_{x\to y}\dfrac1{\frac{f(g(x)) - f(g(y))}{g(x) - g(y)}}= \dfrac1{\lim_{x\to y}\frac{f(g(x)) - f(g(y))}{g(x) - g(y)}}$$
And in this case, you do already know that $f$ is differentiable, and since $g$ is continuous $\lim_{x \to y} g(x) - g(y) = 0$.
