Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 44

Use implicit differentiation to show that $\displaystyle y' = \frac{p}{q} x ^{(p/q) - 1}$ suppose that if $y = x^{p/q}$, then $y^q = x^p$.

Using Power Rule and implicit differentiation.


$\begin{equation} \begin{aligned} y^q =& x^p \\ \\ qy^{q - 1} \frac{dy}{dx} =& px^{p - 1} \\ \\ \frac{dy}{dx} =& \frac{px^{p - 1}}{qy^{q - 1}} \\ \\ \text{but } y =& x^{p/q}, \text{ so } \\ \\ \frac{dy}{dx} =& \frac{px ^{p - 1}}{q [ (x^{p/q})^{q - 1}]} \qquad \qquad \text{ Using the Property of Exponent } \\ \\ \frac{dy}{dx} =& \frac{p}{q} \frac{\displaystyle \left[ \frac{x^p}{x} \right]}{\displaystyle \left[ \frac{\displaystyle \left( \frac{x^{p/q}}{x} \right)^q}{(x^{p/q})} \right]} \\ \\ \frac{dy}{dx} =& \frac{p}{q} \frac{\cancel{x^p} (x^{p/q})}{x \cancel{(x^p)}} \\ \\ \frac{dy}{dx} =& \frac{p}{q} \frac{x^{p/q}}{x} \\ \\ \frac{dy}{dx} =& \frac{p}{q} x^{p/q - 1} \end{aligned} \end{equation}$