How do you use a Maclaurin series to find the derivative of a function?

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How do you use a Maclaurin series to find the derivative of a function?

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Answer 1 · 2014-12-16T05:13:44

The MacLaurin series of a function $f$ $f$ is a power series of the form:

$\infty \sum n = 0 a_{n} x^{n}$ $sum_(n=0)^(oo) a_n x^n$

With the coefficients $a_{n}$ $a_n$ given by the relation

$a_{n} = \frac{f^{(n)} (0)}{n!},$ $a_n=(f^((n))(0))/(n!),$

where $f^{(n)} (0)$ $f^((n))(0)$ is the $n$ $n$ th derivative of $f (x)$ $f(x)$ evaluated at $x = 0$ $x=0$ .

Therefore,

$f^{(n)} (0) = a_{n} n!$ $f^((n))(0)=a_n n!$

This reasoning can be extended to Taylor series around $x_{0}$ $x_0$ , of the form:

$\infty \sum n = 0 c_{n} {(x - x_{0})}_{0}^{n}$ $sum_(n=0)^(oo) c_n (x-x_0)^n$

With the relation

$f^{(n)} (x_{0}) = c_{n} n!$ $f^((n))(x_0)=c_n n!$

It's important to emphasize that the function $n$ $n$ th derivative of $f$ $f$ (that is, $f^{(n)} (x)$ $f^((n)) (x)$ ) cannot be obtained directly from the Taylor/MacLaurin series (only it's value on the point around wich the series is constructed).

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How do you use a Maclaurin series to find the derivative of a function?

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