Answer
See below for detailed proof.
Work Step by Step
$$y^q=x^p$$
1) Differentiate the function implicitly: $$qy^{q-1}\frac{dy}{dx}=px^{p-1}$$ $$\frac{dy}{dx}=\frac{px^{p-1}}{qy^{q-1}}$$ $(y\ne0)$
2) As $y=x^{p/q}$, we substitute $y$ for $x^{p/q}$ here: $$\frac{d}{dx}(x^{p/q})=\frac{px^{p-1}}{q(x^{p/q})^{q-1}}=\frac{px^{p-1}}{qx^{p(q-1)/q}}=\frac{p}{q}x^{p-1-\frac{p(q-1)}{q}}=\frac{p}{q}x^A$$
- Consider $A$: $$A=p-1-\frac{p(q-1)}{q}=\frac{pq-q-p(q-1)}{q}=\frac{pq-q-pq+p}{q}$$ $$A=\frac{p-q}{q}=\frac{p}{q}-1$$
Therefore, $$\frac{d}{dx}(x^{p/q})=\frac{p}{q}x^{(p/q)-1}$$
for $y\ne0$.
