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Theorem:
If $f\,(x) = x^{\frac{p}{q}}$, where $p$ and $q$ are integers, then $f\,'(x) = \frac{p}{q} x^{\frac{p}{q}-1}$
Proof:
Suppose
$$y = x^{\frac{p}{q}}$$Raising both sides to the $q$ power, we get an equation involving integer powers of $x$ and $y$
$$y^q = x^p$$As such, we can differentiate both sides using the Power Rule for Derivatives (for integer powers) to obtain
$$qy^{q-1} \frac{dy}{dx} = px^{p-1}$$Dividing both sides by $qy^{q-1}$ to solve for $\frac{dy}{dx}$, and massaging the result yields
$$\begin{array}{rcl} \frac{dy}{dx} &=& \frac{px^{p-1}}{qy^{q-1}}\\\\ &=& \frac{p}{q} x^{p-1} y^{1-q} \end{array}$$Lastly, we recall that $y = x^{\frac{p}{q}}$, and substitute this expression to remove the $y$ above
$$\begin{array}{rcl} \frac{dy}{dx} &=& \frac{p}{q} x^{p-1} (x^{\frac{p}{q}})^{1-q}\\\\ &=& \frac{p}{q} x^{p-1} x^{\frac{p}{q} - p}\\\\ &=& \frac{p}{q} x^{\frac{p}{q} - 1} \end{array}$$Thus, if $f\,(x) = x^{\frac{p}{q}}$, where $p$ and $q$ are integers, it must be true that $f\,'(x) = \frac{p}{q} x^{\frac{p}{q}-1}$
QED.