Answer
See explanations.
Work Step by Step
Step 1. Differentiating the equivalent equation $y^q = x^p$ implicitly, we can get $qy^{q-1}y'=px^{p-1}$, which gives $y'=\frac{px^{p-1}}{qy^{q-1}}$
Step 2. Using the relation $y=x^{p/q}$, we have $\frac{d}{dx}x^{p/q}=y'=\frac{px^{p-1}}{q(x^{p/q})^{q-1}}=\frac{p}{q}x^{p-1-p(q-1)/q}=\frac{p}{q}x^{(pq-q-pq+p)/q}=\frac{p}{q}x^{(p-q)/q}=\frac{p}{q}x^{p/q-1}$
