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}
$