Polynomials are perhaps the simplest
functions to do
calculus with. Their
derivatives and
integrals are given by the following rules:
- <math>\frac{d}{dx} \sum^n_{r=0} a_r x^r = \sum^n_{r=0} ra_rx^{r-1}</math>
- <math>\int \sum^n_{r=0} a_r x^r\,dx= \sum^n_{r=0} \frac{a_r}{r+1} x^{r+1} + c</math>
Hence the derivative of x^{100} is 100x^{99} and the integral of x^{100} is x^{101}/101 + c.
Proof
Because differentiation is linear, we have:
- <math>\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}</math>
So it remains to find <math>\frac{d\left(x^r\right)}{dx}</math> for any natural number r. The derivative of function f(x) is given by Newton's difference quotient
- <math> f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} </math>
By the binomial theorem, and using the C-notation of combinations,
- <math>\left(x+h\right)^r = \sum_{k=0}^r {}^rC_k h^k x^{r-k}</math>
and therefore
- <math>\frac{\left(x+h\right)^r - x^r}{h} = \sum_{k=1}^{r} {}^rC_k h^{k-1} x^{r-k}</math>
The derivative is the limit of this as <math> h \rightarrow 0 </math>
- <math>\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}</math>
which gives the claimed result.
Generalisation
- <math>\frac{d}{dx} \left(ax^k\right) = akx^{k-1}</math>
is generally true for all values of k where x
^{k} is meaningful. In particular it holds for all rational k for values of x where x
^{k} is defined.
Similarly for integration, see Table of integrals.
References
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