Editing Power rule (section)

====Proof by [[implicit differentiation]]====
A more straightforward generalization of the power rule to rational exponents makes use of implicit differentiation.

Let <math> y=x^r=x^{p/q}</math>, where <math>p, q \in \mathbb{Z}</math> so that <math>r \in \mathbb{Q}</math>.

Then,<math display="block">y^q=x^p</math>Differentiating both sides of the equation with respect to <math>x</math>,<math display="block">qy^{q-1}\cdot\frac{dy}{dx} = px^{p-1}</math>Solving for <math>\frac{dy}{dx}</math>,<math display="block">\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}}.</math>Since <math>y=x^{p/q}</math>,<math display="block">\frac d{dx}x^{p/q} = \frac{px^{p-1}}{qx^{p-p/q}}.</math>Applying laws of exponents,<math display="block">\frac d{dx}x^{p/q} = \frac{p}{q}x^{p-1}x^{-p+p/q} = \frac{p}{q}x^{p/q-1}.</math>Thus, letting <math>r=\frac{p}{q}</math>, we can conclude that <math>\frac d{dx}x^r = rx^{r-1}</math> when <math>r</math> is a rational number.