Power Rule

Type of exponent	Formula	Domain restriction
Positive integer $n$	${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$	all $x$
Negative integer $-n$	${\displaystyle \frac{d}{dx}}{x}^{-n}=-n{x}^{-n-1}$	$x\ne 0$
Rational $m/n$	${\displaystyle \frac{d}{dx}}{x}^{m/n}={\displaystyle \frac{m}{n}}{x}^{m/n-1}$	depends on $n$ even/odd
Any real $r$	${\displaystyle \frac{d}{dx}}{x}^r=r{x}^{r-1}$	(or $x\ne 0$)

The Derivative of $f(x)=x^r$ When $r$ Is a Positive Integer

Consider $f(x)=x^n$, where $n$ is a positive integer.

Read more

To find the derivative, let $x+h=A$ and $x=B$ and use the identity $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +A{B}^{n-2}+B^{n-1})$$ to simplify the difference quotient, then take the limit.

The difference quotient is

$$\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^n-x^n}{h}$$

We know:

$$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +A{B}^{n-2}+B^{n-1})$$

Taking $A=x+h$ and $B=x$:

$$(x+h{)}^n-x^n=h[(x+h{)}^{n-1}+(x+h{)}^{n-2}x+\cdots +x^{n-1}]$$

and

$$\frac{(x+h{)}^n-x^n}{h}=\underset{n\text{ terms}}{\underbrace{(x+h{)}^{n-1}+(x+h{)}^{n-2}x+\cdots +x^{n-1}}}$$

As $h\to 0$, each term approaches $x^{n-1}$, and there are $n$ terms, so the sum approaches $n{x}^{n-1}$.

We conclude: if $f(x)=x^n$ then

f'(x)=nx^{n-1},\qquad n\text{ a positive integer.}

The Derivative of $f(x)=x^r$ When $r$ Is a Negative Integer

When $r=-n$ (), we must assume $x\ne 0$.

Read more

First simplify the difference quotient as $$\frac{{\displaystyle \frac{1}{(x+h{)}^n}}-{\displaystyle \frac{1}{x^n}}}{h}=\frac{x^n-(x+h{)}^n}{h{x}^n(x+h{)}^n}$$ Then let $x=A$ and $x+h=B$ and use the algebraic identity for $A^n-B^n$.

The difference quotient can be written as

\begin{aligned} \frac{\dfrac{1}{(x+h)^n}-\dfrac{1}{x^n}}{h}&=\frac{x^n-(x+h)^n}{hx^n(x+h)^n} \end{aligned}

Setting $A=x$ and $B=x+h$:

\begin{aligned} x^n-(x+h)^n&=-h\big[x^{n-1}+x^{n-2}(x+h)+\cdots+(x+h)^{n-1}\big] \end{aligned}

So

$$\frac{{\displaystyle \frac{1}{(x+h{)}^n}}-{\displaystyle \frac{1}{x^n}}}{h}=-\frac{x^{n-1}+x^{n-2}(x+h)+\cdots +(x+h{)}^{n-1}}{x^n(x+h{)}^n}$$

As $h\to 0$ each of the $n$ numerator terms approaches $x^{n-1}$, giving

$$\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=-\frac{n{x}^{n-1}}{x^{2n}}=-n{x}^{-n-1}=r{x}^{r-1}$$

If $r$ is a negative integer:

f(x)=x^r \Rightarrow f'(x)=rx^{r-1}

The Derivative of $f(x)=x^r$ When $r$ Is a Rational Number

Suppose $r=m/n$ where $m$ and $n$ are integers.

Read more

Let $(x+h{)}^{1/n}=A$ and $x^{1/n}=B$ to write the difference quotient as ${\displaystyle \frac{A^m-B^m}{A^n-B^n}}$. Then use the identity $A^N-B^N=(A-B)(A^{N-1}+\cdots +B^{N-1})$ for both numerator and denominator.

We have

$$\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^{m/n}-x^{m/n}}{h}$$

Let $A=(x+h{)}^{1/n}$ and $B=x^{1/n}$, so $h=A^n-B^n$. The difference quotient becomes

\begin{aligned} \frac{f(x+h)-f(x)}{h}&=\frac{A^m-B^m}{A^n-B^n}\\ &=\frac{(A-B)(\overbrace{A^{m-1}+A^{m-2}B+\cdots+B^{m-1}}^{m\text{ terms}})}{(A-B)(\underbrace{A^{n-1}+A^{n-2}B+\cdots+B^{n-1}}_{n\text{ terms}})} \end{aligned}

As $h\to 0$ (equivalently $A\to B$):

\begin{aligned} \lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=\frac{mB^{m-1}}{nB^{n-1}}=\frac{m}{n}B^{m-n}=\frac{m}{n}x^{m/n-1} \end{aligned}

If $r$ is a rational number:

f(x)=x^r \Rightarrow f'(x)=rx^{r-1}

The General Case: Any Real Exponent

For $x\ne 0$ and any real $r$, write the difference quotient as

\begin{aligned} \frac{f(x+h)-f(x)}{h}&=\frac{(x+h)^r-x^r}{h}\\ &=x^{r-1}\frac{\left(1+\dfrac{h}{x}\right)^r-1}{\dfrac{h}{x}} \end{aligned}

Setting $u=h/x$ and using the previously proved limit ${\displaystyle \underset{u\to 0}{lim}\frac{(1+u{)}^r-1}{u}=r}$:

\begin{aligned} \lim_{h\to0}\frac{f(x+h)-f(x)}{h}&=x^{r-1}\times r=rx^{r-1} \end{aligned}

For $x=0$ and $r\ge 1$ the definition of the derivative gives f'(0)=0.

Frequently Asked Questions

Does the Power Rule apply to irrational exponents like $x^{\sqrt{2}}$?

Yes. The general Power Rule holds for any real exponent $r$. The proof for irrational $r$ uses logarithmic differentiation (covered in Section 3.11) or the limit $\underset{u\to 0}{lim}(1+u{)}^r/u=r$.

What is ${\displaystyle \frac{d}{dx}}\sqrt{x}$?

Write $\sqrt{x}=x^{1/2}$ and apply the Power Rule: ${\displaystyle \frac{d}{dx}}{x}^{1/2}={\displaystyle \frac{1}{2}}{x}^{-1/2}={\displaystyle \frac{1}{2\sqrt{x}}}$.

What about ${\displaystyle \frac{d}{dx}}(1/x^3)$?

Write $1/x^3=x^{-3}$ and apply the Power Rule: ${\displaystyle \frac{d}{dx}}{x}^{-3}=-3x^{-4}=-{\displaystyle \frac{3}{x^4}}$.