Power rule proof

Question

I've already proven that $\frac{d}{dx}(x^n)=nx^{n-1}$ for all $n\in \mathbb{N}$ using induction and the product rule, and that $\frac{d}{dx}(x^{-n})=-nx^{-n-1}$ for all $n\in \mathbb{N}$ using the quotient rule.

Now I am asked to prove $\frac{d}{dx}(x^{1/n})=\frac{1}{n}x^{\frac{1}{n}-1}$ for all $n\in \mathbb{Z} \setminus \{0\}$. And finally, to prove $\frac{d}{dx}(x^{\frac{p}{q}})=\frac{p}{q}x^{\frac{p}{q}-1}$ for all $\frac{p}{q}\in \mathbb{Q}$. According to the hints, the chain rule is useful, but I don't entirely understand how to apply the chain rule here. Could someone clarify this? Also, should I be using induction to solve these problems? I don't necessarily see how to apply induction to the first question since we are interested in $n\in \mathbb{Z}$. Here is the "hint" for the first proof: If $y(x)=x^{\frac{1}{n}}=x^{\frac{1}{n}}$ then $(y(x))^n=x$. Now use the chain rule to find an expression that contains $\frac{dy}{dx}$ and isolate $\frac{dy}{dx}$ to be by itself on one side of the expression. It's unclear to me how to apply $\frac{dy}{dx}$ in this situation.

Thanks a lot.

Answer 1

Note that derivating both side and using chain rule for LHS we have

$$(y(x))^n=x\implies n\cdot (y(x))^{n-1}\cdot y'(x)=1\implies y'(x)=\frac1n(y(x))^{1-n}=\frac1n x^{\frac1n-1}$$

and similarly for $x^\frac{p}{q}$.

Answer 2

For the first case you have $$ n y^{n-1}\frac{dy}{dx}=1\iff\frac{dy}{dx}=\frac{1}{n\cdot y^{n-1}} $$ and using $y=x^{1/n}$ should get you there.

The other case is more straight forward, since it is just an application of the chain rule with $y(x)=x^{1/q}$ and considering $(y(x))^p=x^{p/q}$.