Power Rule for Derivatives

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

$\map {\dfrac \d {\d x} } {c x^n} = n c x^{n - 1}$

Proof

This can be done in sections.

Proof for Natural Number Index

Let $\map f x = x^n$ for $x \in \R, n \in \N$.

By the definition of the derivative:

$\ds \dfrac \d {\d x} \map f x = \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h = \lim_{h \mathop \to 0} \dfrac {\paren {x + h}^n - x^n} h$

Using the binomial theorem this simplifies to:

 $\ds$  $\ds \lim_{h \mathop \to 0} \paren {\frac {\dbinom n 0 x^n + \dbinom n 1 x^{n - 1} h + \dbinom n 2 x^{n - 2} h^2 + \cdots + \dbinom n {n - 1} x h^{n - 1} + \dbinom n n h^n - x^n} h}$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\frac {\dbinom n 1 x^{n - 1} h + \dbinom n 2 x^{n - 2} h^2 + \cdots + \dbinom n {n - 1} x h^{n - 1} + \dbinom n n h^n} h}$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \paren {\dbinom n 1 x^{n - 1} + \dbinom n 2 x^{n - 2} h^1 + \cdots + \dbinom n {n - 1} x h^{n - 2} + \dbinom n n h^{n - 1} }$ $\ds$ $=$ $\ds \dbinom n 1 x^{n - 1}$ evaluating the limit $\ds$ $=$ $\ds n x^{n - 1}$ 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 $x^n = \dfrac 1 {x^m}$.

 $\ds \map D {x^n}$ $=$ $\ds \map D {\frac 1 {x^m} }$ $\ds$ $=$ $\ds \frac {x^m \cdot 0 - 1 \cdot m x^{m - 1} } {x^{2 m} }$ Quotient Rule for Derivatives $\ds$ $=$ $\ds -m x^{-m - 1}$ $\ds$ $=$ $\ds n x^{n - 1}$

$\blacksquare$

Proof for Fractional Index

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

Thus, let $\map f x = x^{1 / n}$.

From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $\map f x$ is defined when $x \ge 0$.

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

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

Let $y > x$.

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

where $x, y \in \closedint X Y$.

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

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

From the Squeeze Theorem, it follows that:

$\ds \lim_{y \mathop \to x^+} \dfrac {y^{1 / n} - x^{1 / n} } {y - x} = \dfrac 1 {n x} x^{1 / n} = \dfrac 1 n x^{\dfrac 1 n - 1}$

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

Thus the result holds for $\map f x = x^{1 / n}$.

$\blacksquare$

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:

 $\ds \map D {x^n}$ $=$ $\ds \map D {x^{p/q} }$ $\ds$ $=$ $\ds \map D {\paren {x^p}^{1/q} }$ $\ds$ $=$ $\ds \frac 1 q \paren {x^p}^{1/q} x^{-p} p x^{p-1}$ Chain Rule for Derivatives $\ds$ $=$ $\ds \frac p q x^{\frac p q - 1}$ after some algebra $\ds$ $=$ $\ds n x^{n - 1}$

$\blacksquare$

Proof for Real Number Index

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

To do this, we compute the limit $\ds \lim_{h \mathop \to 0} \frac {\paren {x + h}^n - x^n} h$:

 $\ds \frac {\paren {x + h}^n - x^n} h$ $=$ $\ds \frac {x^n} h \paren {\paren {1 + \frac h x}^n - 1}$ $\ds$ $=$ $\ds \frac{x^n} h \paren {e^{n \map \ln {1 + \frac h x} } - 1}$ $\ds$ $=$ $\ds x^n \cdot \frac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \frac h x} } \cdot \frac {n \map \ln {1 + \frac h x} } {\frac h x} \cdot \frac 1 x$

Now we use the following results:

$\ds \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponential at Zero
$\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$ from Derivative of Logarithm at One

to obtain:

$x^n \cdot \dfrac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \dfrac h x} } \cdot \dfrac {n \map \ln {1 + \dfrac h x}} {\dfrac h x} \cdot \dfrac 1 x \to n x^{n - 1}$ 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 $x^m$.