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.

Proof
This can be done in sections.

Proof for Natural Number Index
Let $$f(x)=x^n$$,$$x\in R$$ and $$n\in N$$.

By the definition of the derivative, $$\frac{d}{dx}f(x) = \lim_{h \to 0} \frac{f (x+h) - f(x)} {h} = \lim_{h \to 0} \frac{(x+h)^n-x^n}{h}$$.

Using the binomial theorem this simplifies to:

$$ $$ $$ $$ $$

Alternative Proof for Natural Number Index
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.

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

$$ $$ $$ $$

Proof for Fractional Index
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:


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

From the Squeeze Theorem, it follows that $$\lim_{y \to x^+} \frac {y^{1/n} - x^{1/n}} {y - x} = \frac 1 {n x} x^{1/n}= \frac 1 n x^{\frac 1 n - 1}$$.

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

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

Alternative Proof for Fractional Index
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:

$$ $$ $$ $$

Proof for Rational Index
Let $$n \in \Q$$, such that $$n = \frac p q$$ where $$p, q \in \Z, q \ne 0$$.

Then we have:

$$ $$ $$ $$ $$

Proof for Real Number Index
We are going to prove that $$f^{\,'}(x) = n x^{n-1}$$ holds for all real $$n$$.

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

$$ $$ $$

Now we use the following results:
 * $$\lim_{x \to 0} \frac {\exp x - 1} {x} = 1$$ from Derivative of Exponent at Zero;
 * $$\lim_{x \to 0} \frac {\ln \left({1 + x}\right)} {x} = 1$$ from Derivative of Logarithm at One.

... to obtain:
 * $$\frac {e^{n \ln \left({1 + \frac{h}{x}}\right)} - 1} {n \ln \left( {1 + \frac{h}{x}}\right)} \cdot \frac {n \ln \left({1 + \frac{h}{x}}\right)} {\frac{h}{x}} \cdot \frac{1}{x} \to n x^{n-1}$$

as $$h \to 0$$.

Hence the result.