Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 48

$y' = \frac{p}{q}~x^{(p/q)-1}$

$y^q = x^p$ We can find an expression for $y'$: $qy^{q-1}~y' = px^{p-1}$ $y' = \frac{p}{q}~\frac{x^{p-1}}{y^{q-1}}$ $y' = \frac{p}{q}~\frac{x^{p-1}}{y^q/y}$ $y' = \frac{p}{q}~\frac{(x^{p-1})~(y)}{y^q}$ $y' = \frac{p}{q}~\frac{(x^{p-1})~(x^{p/q})}{x^p}$ $y' = \frac{p}{q}~(x^{-1})~(x^{p/q})$ $y' = \frac{p}{q}~x^{(p/q)-1}$