Calculus with polynomials

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In mathematics, polynomials are perhaps the simplest functions with which to do calculus. Their derivatives and indefinite integrals are given by the following rules:

[\left( \sum^n_ a_k x^k\right)' = \sum^n_ ka_kx^]

and

[\int\!\left( \sum^n_ a_k x^k\right)\,dx= \sum^n_ \frac} + c.]

Hence, the derivative of [x^] is 100x⁹⁹ and theindefinite integral of [x^] is [x^/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.

Contents

1 The power rule
2 Proof of the power rule
3 Differentiation of arbitrary polynomials
4 Generalization
5 See also
6 References

The 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^, ] that is,

[\left(x^n\right)'=nx^.]

The power rule for integration

[\int\! x^n \, dx=\frac}+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.

Proof of the power rule

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

[f'(x) = \lim_ \frac. ]

Substituting [ f(x) = x^n ] gives

[f'(x) = \lim_ \frac.]

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

[f'(x) = \lim_ \frac^ -x^n}. ]

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

[f'(x) = \lim_ \frac^ + x^n -x^n}. ]

Canceling the [x^n] terms one generates

[f'(x) = \lim_ \frac^ }.]

An [h] can be factored out from each term in the sum to give

[f'(x) = \lim_ \frac^ }.]

From thence we can cancel the [h] in the denominator to obtain

[f'(x) = \lim_ \sum_^ .]

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) = } x^. ]

Evaluating the binomial coefficient gives

[} = \frac = \frac = n.]

It follows that

[f'(x) = n x^. \! ]

Differentiation of arbitrary polynomials

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

[\left( \sum_^n a_r x^r \right)' =\sum_^n \left(a_r x^r\right)' =\sum_^n a_r \left(x^r\right)' =\sum_^n ra_rx^.]

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

[\int\!\left( \sum^n_ a_k x^k\right)\,dx= \sum^n_ \frac} + c.]

Generalization

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

[\left(x^a\right)' = ax^]

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^\, 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 061822307X.