Power rule

From Wikipedia, the free encyclopedia
  (Redirected from Derivative of a constant)
Jump to: navigation, search

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 separate 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[edit]

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,


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.


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[edit]

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.


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

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


\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.


  • 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.