# Power rule

In calculus, the power rule is one of the most important differentiation rules. Since differentiation is linear, polynomials can be differentiated using this rule.

$\frac{d}{dx} x^n = nx^{n-1} , \qquad n \neq 0.$

The power rule holds for all powers except for the constant value $x^0$ which is covered by the constant rule. The derivative is just $0$ rather than $0 \cdot x^{-1}$ which is undefined when $x=0$.

The inverse of the power rule enables all powers of a variable $x$ except $x^{-1}$ to be integrated. This integral is called Cavalieri's quadrature formula and was first found in a geometric form by Bonaventura Cavalieri for $n \ge 0$. It is considered the first general theorem of calculus to be discovered.

$\int\! x^n \,dx= \frac{ x^{n+1}}{n+1} + C, \qquad n \neq -1.$

This is an indefinite integral where $C$ is the arbitrary constant of integration.

The integration of $x^{-1}$ requires a rule.

$\int \! x^{-1}\, dx= \ln |x|+C,$

Hence, the derivative of $x^{100}$ is $100 x^{99}$ and the integral of $x^{100}$ is $\frac{1}{101} x^{101} +C$.

## Power rule

Historically the power rule was derived as the inverse of Cavalieri's quadrature formula which gave the area under $x^n$ for any integer $n \geq 0$. Nowadays the power rule is derived first and integration considered as its inverse.

For integers $n \geq 1$, the derivative of $f(x)=x^n \!$ is $f'(x)=nx^{n-1},\!$ that is,

$\left(x^n\right)'=nx^{n-1}.$

The power rule for integration

$\int\! x^n \, dx=\frac{x^{n+1}}{n+1}+C$

for $n \geq 0$ is then an easy consequence. One just needs to take the derivative of this equality and use the power rule and linearity of differentiation on the right-hand side.

### Proof

To prove the power rule for differentiation, we use the definition of the derivative as a limit. But first, note the factorization for $n \geq 1$:

$f(x)-f(a) = x^n-a^n = (x-a)(x^{n-1}+ax^{n-2}+ \cdots +a^{n-2}x+a^{n-1})$

Using this, we can see that

$f'(a) = \lim_{x\rarr a} \frac{x^n-a^n}{x-a} = \lim_{x\rarr a} x^{n-1}+ax^{n-2}+ \cdots +a^{n-2}x+a^{n-1}$

Since the division has been eliminated and we have a continuous function, we can freely substitute to find the limit:

$f'(a) = \lim_{x\rarr a} x^{n-1}+ax^{n-2}+ \cdots +a^{n-2}x+a^{n-1} = a^{n-1}+a^{n-1}+ \cdots +a^{n-1}+a^{n-1} = n\cdot a^{n-1}$

The use of the quotient rule allows the extension of this rule for n as a negative integer, and the use of the laws of exponents and the chain rule allows this rule to be extended to all rational values of $n$ . For an irrational $n$, a rational approximation is appropriate.

## Differentiation of arbitrary polynomials

To differentiate arbitrary polynomials, one can use the linearity property of the differential operator to obtain:

$\left( \sum_{r=0}^n a_r x^r \right)' = \sum_{r=0}^n \left(a_r x^r\right)' = \sum_{r=0}^n a_r \left(x^r\right)' = \sum_{r=0}^n ra_rx^{r-1}.$

Using the linearity of integration and the power rule for integration, one shows in the same way that

$\int\!\left( \sum^n_{k=0} a_k x^k\right)\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1} + C.$

## Generalizations

One can prove that the power rule is valid for any exponent r, that is

$\frac{d}{dx}x^r = rx^{r-1}$

Proof:

$\ln x^r=r\ln x$
$\frac{d}{dx} \ln x^r = \frac{d}{dx}(r \ln x)$
$\frac{1}{x^r} \cdot \frac{d}{dx}x^r = r \cdot \frac{d}{dx}\ln x$ (chain rule and constant factor rule)
$\frac{1}{x^r}\cdot \frac{d}{dx}x^r = \frac{r}{x}$
$\frac{d}{dx}x^r = \frac{rx^r}{x}$
$\frac{d}{dx}x^r = rx^{r-1}$

as long as x is in the domain of the functions on the left and right hand sides and r is nonzero. Using this formula, together with

$\int \! x^{-1}\, dx= \ln |x|+C,$

one can differentiate and integrate linear combinations of powers of x which are not necessarily polynomials.

## References

• Larson, Ron; Hostetler, Robert P.; and Edwards, Bruce H. (2003). Calculus of a Single Variable: Early Transcendental Functions (3rd edition). Houghton Mifflin Company. ISBN 0-618-22307-X.