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In mathematics, polynomials are perhaps the simplest functions used in calculus. Their derivatives and indefinite integrals are given by the following rules:

$\left( \sum^n_{k=0} a_k x^k\right)' = \sum^n_{k=1} ka_kx^{k-1}$

and

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

Hence, the derivative of $x 100$ is $100 x 99$ and the indefinite integral of $x 100$ is $\frac{x^{101}}{101}+C$ where C is an arbitrary constant of integration.

This article will state and prove the power rule for differentiation, and then use it to prove these two formulas.

[edit] Power rule

The power rule for differentiation states that for every natural number n, 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 natural n 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.

[edit] Proof

To prove the power rule for differentiation, we use the definition of the derivative as a limit:

$f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}.$

Substituting $f (x) = x n$ gives

$f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.$

One can then express $(x + h) n$ by applying the binomial theorem to obtain

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n} {{n \choose i} x^i h^{n-i}}-x^n}{h}.$

The $i = n$ term of the sum can then be written independently of the sum to yield

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}} + x^n -x^n}{h}.$

Cancelling the $x n$ terms one generates

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}}}{h}.$

An $h$ can be factored out from each term in the sum. From thence we can cancel the h in the denominator to obtain

$f'(x) = \lim_{h\rarr0} \sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i-1}}.$

To evaluate this limit we observe that $n - i - 1 > 0$ for all $i < n - 1$ and equal to zero for $i=n-1.\,\!$ Thus only the $h 0$ term will survive with $i = n - 1$ yielding

$f'(x) = {n \choose {n-1}} x^{n-1}.$

Evaluating the binomial coefficient gives

${n \choose {n-1}} = \frac{n!}{(n-1)!\ 1!} = \frac{n\ (n-1)!}{(n-1)!} = n.$

It follows that

$f'(x) = n x^{n-1}. \!$

[edit] 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

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

[edit] Generalization

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

$\left(x^a\right)' = ax^{a-1}$

for any real number a as long as x is in the domain of the functions on the left and right hand sides. 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.

[edit] 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.

Contents

[edit] Power rule

[edit] Proof

[edit] Differentiation of arbitrary polynomials

[edit] Generalization

[edit] References