## Encyclopedia > Calculus with polynomials

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# Calculus with polynomials

Polynomials are perhaps the simplest functions to do calculus with. Their derivatives and integrals are given by the following rules:

$\frac{d}{dx} \sum^n_{r=0} a_r x^r = \sum^n_{r=0} ra_rx^{r-1}$
$\int \sum^n_{r=0} a_r x^r\,dx= \sum^n_{r=0} \frac{a_r}{r+1} x^{r+1} + c$

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

Proof Because differentiation is linear, we have:

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

So it remains to find $\frac{d\left(x^r\right)}{dx}$ for any natural number r. The derivative of function f(x) is given by Newton's difference quotient

$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$

By the binomial theorem, and using the C-notation of combinations,

$\left(x+h\right)^r = \sum_{k=0}^r {}^rC_k h^k x^{r-k}$

and therefore

$\frac{\left(x+h\right)^r - x^r}{h} = \sum_{k=1}^{r} {}^rC_k h^{k-1} x^{r-k}$

The derivative is the limit of this as $h \rightarrow 0$

$\frac{d}{dx}\left(x^r\right) = \lim_{h\rightarrow 0} \left(\sum_{k=1}^{r} {}^rC_k h^{k-1} x^{r-k}\right) = {}^rC_1 x^{r-1} = rx^{r-1}$

which gives the claimed result. Generalisation

$\frac{d}{dx} \left(ax^k\right) = akx^{k-1}$
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.

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