Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises - Page 167: 48

See proof

$$qy^{q-1}y'=px^{p-1}$$ $$q\frac{y^{q}{y}}y'=p\frac{x^{p}}{x}$$ $$q\frac{x^{p}{y}}y'=p\frac{x^{p}}{x}$$ $$q\frac{{y}}y'=p\frac{1}{x}$$ $$y'=\frac{p}{q}\frac{y}{x}$$ $$y'=\frac{p}{q}\frac{x^{\frac{p}{q}}}{x}$$ $$y'=\frac{p}{q}x^{\frac{p}{q}-1}$$