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===Generalization to rational exponents=== Upon proving that the power rule holds for integer exponents, the rule can be extended to rational exponents. ====Proof by [[chain rule]]==== This proof is composed of two steps that involve the use of the chain rule for differentiation. # Let <math>y=x^r=x^\frac1n</math>, where <math>n\in\N^+</math>. Then <math>y^n=x</math>. By the [[chain rule]], <math>ny^{n-1}\cdot\frac{dy}{dx}=1</math>. Solving for <math>\frac{dy}{dx}</math>, <math display="block">\frac{dy}{dx} =\frac{1}{ny^{n-1}} =\frac{1}{n\left(x^\frac1n\right)^{n-1}} =\frac{1}{nx^{1-\frac1n}} =\frac{1}{n}x^{\frac1n-1} =rx^{r-1}</math>Thus, the power rule applies for rational exponents of the form <math>1/n</math>, where <math>n</math> is a nonzero natural number. This can be generalized to rational exponents of the form <math>p/q</math> by applying the power rule for integer exponents using the chain rule, as shown in the next step. # Let <math>y=x^r=x^{p/q}</math>, where <math>p\in\Z, q\in\N^+,</math> so that <math>r\in\Q</math>. By the [[chain rule]], <math display="block">\frac{dy}{dx} =\frac{d}{dx}\left(x^\frac1q\right)^p =p\left(x^\frac1q\right)^{p-1}\cdot\frac{1}{q}x^{\frac1q-1} =\frac{p}{q}x^{p/q-1}=rx^{r-1}</math> From the above results, we can conclude that when <math>r</math> is a [[rational number]], <math>\frac{d}{dx} x^r=rx^{r-1}.</math> ====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.
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