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
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:

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

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:


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

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

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

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

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 $\displaystyle n = \frac p q$ where $p, q \in \Z, q \ne 0$.

Then we have:

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

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

Now we use the following results:
 * $\displaystyle \lim_{x \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponent 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.

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