Power Rule for Derivatives

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Theorem

Let $n \in \R$.

Let $f: \R \to \R$ be the real function defined as $\map f x = x^n$.


Then:

$\map {f'} x = n x^{n - 1}$

everywhere that $\map f x = x^n$ is defined.


When $x = 0$ and $n = 0$, $\map {f'} x$ is undefined.


Corollary

$\dfrac \d {\d x} \left({c x^n}\right) = n c x^{n - 1}$


Proof

This can be done in sections.

Proof for Natural Number Index

Let $f \left({x}\right) = x^n$ for $x \in \R, n \in \N$.

By the definition of the derivative:

$\displaystyle \dfrac \d {\d x} f \left({x}\right) = \lim_{h \mathop \to 0} \dfrac {f \left({x + h}\right) - f \left({x}\right)} h = \lim_{h \mathop \to 0} \dfrac{(x + h)^n - x^n} h$


Using the binomial theorem this simplifies to:

\(\displaystyle \) \(\) \(\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 0} x^n + {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n - x^n} h}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \left({\frac { {n \choose 1} x^{n - 1} h + {n \choose 2} x^{n - 2} h^2 + \cdots + {n \choose n - 1} x h^{n - 1} + {n \choose n} h^n} h}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \left({ {n \choose 1} x^{n - 1} + {n \choose 2} x^{n - 2} h^1 + \cdots + {n \choose n - 1} x h^{n - 2} + {n \choose n} h^{n - 1} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle {n \choose 1} x^{n - 1}\) evaluating the limit
\(\displaystyle \) \(=\) \(\displaystyle n x^{n - 1}\) Binomial Coefficient with One: $\dbinom r 1 = r$

$\blacksquare$


Proof for Integer Index

When $n \ge 0$ we use the result for Natural Number Index.

Now let $n \in \Z: n < 0$.

Then let $m = -n$ and so $m > 0$.

Thus $x^n = \dfrac 1 {x^m}$.

\(\displaystyle D \left({x^n}\right)\) \(=\) \(\displaystyle D \left({\frac 1 x^m}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^m \cdot 0 - 1 \cdot m x^{m-1} } {x^{2 m} }\) Quotient Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle -m x^{-m-1}\)
\(\displaystyle \) \(=\) \(\displaystyle n x^{n-1}\)

$\blacksquare$


Proof for Fractional Index

Let $n \in \N_{>0}$.

Thus, let $f \left({x}\right) = x^{1/n}$.

From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $f \left({x}\right)$ is defined when $x \ge 0$.

(However, see the special case where $x = 0$.)

From Continuity of Root Function, $f \left({x}\right)$ is continuous over the open interval $\left({0 \,.\,.\, \infty}\right)$, but not at $x = 0$ where it is continuous only on the right.


Let $y > x$.

From Inequalities Concerning Roots:

$\forall n \in \N_{>0}: X Y^{1/n} \ \left|{x - y}\right| \le n X Y \ \left|{x^{1/n} - y^{1/n}}\right| \le Y X^{1/n} \ \left|{x - y}\right|$

where $x, y \in \left[{X \,.\,.\, Y}\right]$.

Setting $X = x$ and $Y = y$, this reduces (after algebra) to:

$\displaystyle \frac 1 {n y} y^{1/n} \le \frac {y^{1/n} - x^{1/n}} {y - x} \le \frac 1 {n x} x^{1/n}$

From the Squeeze Theorem, it follows that:

$\displaystyle \lim_{y \to x^+} \ \frac {y^{1/n} - x^{1/n}} {y - x} = \frac 1 {n x} x^{1/n} = \frac 1 n x^{\frac 1 n - 1}$


A similar argument shows that the left hand limit is the same.


Thus the result holds for $f \left({x}\right) = x^{1/n}$.

$\blacksquare$


Proof for Rational Index

Let $n \in \Q$, such that $n = \dfrac p q$ where $p, q \in \Z, q \ne 0$.

Then we have:


\(\displaystyle D \left({x^n}\right)\) \(=\) \(\displaystyle D \left({x^{p/q} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle D \left({\left({x^p}\right)^{1/q} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 q \left({x^p}\right)^{1/q} x^{-p} p x^{p-1}\) Chain Rule
\(\displaystyle \) \(=\) \(\displaystyle \frac p q x^{\frac p q - 1}\) after some algebra
\(\displaystyle \) \(=\) \(\displaystyle n x^{n - 1}\)

$\blacksquare$


Proof for Real Number Index

We are going to prove that $f' \left({x}\right) = n x^{n-1}$ holds for all real $n$.

To do this, we compute the limit $\displaystyle \lim_{h \to 0} \frac{\left({x + h}\right)^n - x^n} h$:


\(\displaystyle \frac{\left({x + h}\right)^n - x^n} h\) \(=\) \(\displaystyle \frac{x^n} h \left({\left({1 + \frac h x}\right)^n - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac{x^n} h \left({e^{n \ln \left({1 + \frac h x}\right)} - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle x^n \cdot \frac {e^{n \ln \left({1 + \frac h x}\right)} - 1} {n \ln \left({1 + \frac h x}\right)} \cdot \frac {n \ln \left({1 + \frac h x}\right)} {\frac h x} \cdot \frac 1 x\)


Now we use the following results:

$\displaystyle \lim_{x \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponential at Zero
$\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} x = 1$ from Derivative of Logarithm at One

... to obtain:

$x^n \cdot \dfrac {e^{n \ln \left({1 + \frac h x}\right)} - 1} {n \ln \left( {1 + \dfrac h x}\right)} \cdot \dfrac {n \ln \left({1 + \dfrac h x}\right)} {\dfrac h x} \cdot \dfrac 1 x \to n x^{n-1}$ as $h \to 0$

Hence the result.

$\blacksquare$


Historical Note

The Power Rule for Derivatives was stated, without proof or explanation, by Gottfried Wilhelm von Leibniz in his $1684$ article Nova Methodus pro Maximis et Minimis, published in Acta Eruditorum.

Isaac Newton had established exactly the same result in a privately circulated paper of $1669$: On Analysis by Means of Equations with an Infinite Number of Terms, by investigation the nature of a function whose area under the graph is $x^m$.


Sources