# Calculus with polynomials

 Topics in calculus Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials | Mean value theorem | Vector calculus | Tensor calculus Differentiation Integration

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

[itex]\frac{d}{dx} \sum^n_{k=0} a_k x^k = \sum^n_{k=0} ka_kx^{k-1}[itex]
[itex]\int \sum^n_{k=0} a_k x^k\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1} + c.[itex]

Hence the derivative of x100 is 100x99 and the integral of x100 is x101/101 + c.

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## Proof

Because differentiation is linear, we have:

[itex]\frac{d\left( \sum_{r=0}^n a_r x^r \right)}{dx} =

\sum_{r=0}^n \frac{d\left(a_r x^r\right)}{dx} = \sum_{r=0}^n a_r \frac{d\left(x^r\right)}{dx}.[itex]

So it remains to find [itex]\frac{d\left(x^r\right)}{dx}[itex] for any natural number r. There is a proof by induction using the product rule; this depends only on the case r = 1.

## Generalisations

[itex]\frac{d}{dx} \left(ax^k\right) = akx^{k-1}[itex]

is generally true for all values of k where xk is meaningful. In particular it holds for all rational k for values of x where xk is defined.

Similarly for integration, see table of integrals.

If one has polynomials with coefficients that are not real or complex numbers (perhaps they are integers, or numbers modulo a prime number) then one can formally define the derivative according to the rules given above. This is useful, for example, in determining whether a polynomial will have multiple roots: compute the greatest common divisor of the polynomial and its formal derivative. If this polynomial is zero, then the original polynomial cannot have any multiple roots.

## 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.

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