Power Rule for Derivatives

Theorem

Let $n \in \R$.

Let $f: \R \to \R$ be the real function defined as $f \left({x}\right) = x^n$.

Then:

$f^{\prime} \left({x}\right) = n x^{n-1}$

everywhere that $f \left({x}\right) = x^n$ is defined.

When $x = 0$ and $n = 0$, $f^{\prime} \left({x}\right)$ is undefined.

Corollary

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

Proof

This can be done in sections.

Proof for Natural Number Index

Proof by Binomial Theorem

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

By the definition of the derivative:

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

Using the binomial theorem this simplifies to:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle \lim_{h \to 0} \left({\frac{ {n \choose 0} x^n + {n \choose 1} x^{n-1} h + {n \choose 2} x^{n-2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n - x^n} h}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \left({\frac{ {n \choose 1} x^{n-1} h + {n \choose 2} x^{n-2} h^2 + \cdots + {n \choose n-1} x h^{n-1} + {n \choose n} h^n} h}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \left({ {n \choose 1} x^{n-1} + {n \choose 2} x^{n-2} h^1 + \cdots + {n \choose n-1} x h^{n-2} + {n \choose n} h^{n-1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle {n \choose 1} x^{n-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	evaluating the limit
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n x^{n-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	value of binomial coefficient: $\displaystyle \binom r 1 = r$

$\blacksquare$

Proof by Induction

We will use the notation $D f \left({x}\right) = f^{\prime} \left({x}\right)$ as it is convenient.

Let $n = 0$.

Then $\forall x \in \R: x^n = 1$.

Thus $f \left({x}\right)$ is the constant function $f_1 \left({x}\right)$ on $\R$.

Thus from Differentiation of a Constant, $D f \left({x}\right) = D \left({x^0}\right) = 0 x^{-1}$, except where $x = 0$.

So the result holds for $n = 0$.

Let $n = 1$.

Then $\forall x \in \R: f \left({x}\right) = x^n = x$.

Then from Differentiation of the Identity Function $D \left({x}\right) = 1 = 1 \cdot x^{1-1}$.

So the result holds for $n = 1$.

Now assume $D \left({x^k}\right) = k x^{k-1}$.

Then by the Product Rule for Derivatives, $D \left({x^{k+1}}\right) = D \left({x^k x}\right) = x^k D \left({x}\right) + D \left({x^k}\right) x = x^k \cdot 1 + k x^{k-1} x = \left({k+1}\right) x^k$.

The result follows by induction.

$\blacksquare$

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 $\displaystyle x^n = \frac 1 {x^m}$.

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

$\blacksquare$

Proof for Fractional Index

Proof from First Principles

Let $n \in \N^*$.

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

From the definition of power for rational numbers, 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, we have:

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

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 using Derivative of Inverse

Let $n \in \N^*$.

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

Thus $f^{-1} \left({y}\right) = x = y^n$ from the definition of root.

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D x^{1/n}\)	\(=\)	\(\displaystyle \frac 1 {D y^n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of an Inverse Function‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {n y^{n-1} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Power Rule for Derivatives: Integer Index
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {n \left({x^{1/n} }\right)^{n-1} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 n x^{\frac 1 n - 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof for Rational Index

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

Then we have:

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

$\blacksquare$

Proof for Real Number Index

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

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

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

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:

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

Hence the result.

$\blacksquare$

Proof of Corollary

Follows directly from the above, and Derivative of Constant Multiple.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.4$