  ## Proof - Power Rule for Rational Powers

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.