# Proving the “differentiablity” of $x^{1/n}$

I am solving a problem that is below

Let $n \in \mathbb{N}$ and let $f(x) = x^{1/n}$, $x \gt 0$. Prove that $f$ is differentiable on $(0,\infty)$ and find $f\,'$. (Hint: Note that $f = g^{−1}$, where $g(y) = y^n$, $y \gt 0$.)

My question is

1) I should use induction correct?

2) I am not sure how to get $f\,'$ via the hint. I know that $(f^{-1})'=1/f'(f^{-1})$ though I am not sure how to apply this...

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The Inverse Function Theorem says that if $g(x)$ is differentiable at $a$, and $g'(a)\neq 0$, then $g^{-1}$ is differentiable at $g(a)$ and $$(g^{-1})'(g(a)) = \frac{1}{g'(a)}.$$
Since $f(x) = x^{1/n}$, in order to show that $f(x)$ is differentiable at $a^n$ it is enough to show that $g(x)=x^n$ is differentiable at $a$ and that $g'(a)\neq 0$, by the Inverse Function Theorem, since $f(x) = g^{-1}(x)$.
Now, if you know that $g(x)=x^n$ is differentiable for all $n$, then you don't need to use induction, just use the Inverse Function Theorem, noting that every value in $(0,\infty)$ is the image of an $a$ in $(0,\infty)$ under $g$, and that $g'(a)\neq 0$ if $a\neq 0$.
If you don't yet know that $g(x)=x^n$ is differentiable, then you can use induction: $g(x)=x$ is differentiable; if $h(x)=x^n$ is differentiable, then $g(x)=x^{n+1} = h(x)x$, and the product of differentiable functions is differentiable, hence $g(x)$ is differentiable. By induction, $x\mapsto x^n$ is differentiable for all $n\in\mathbb{N}$.