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

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

${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$

With the coefficients ${a}_{n}$ given by the relation

a_n=(f^((n))(0))/(n!),

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

Therefore,

f^((n))(0)=a_n n!

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

${\sum}_{n = 0}^{\infty} {c}_{n} {\left(x - {x}_{0}\right)}^{n}$

With the relation

f^((n))(x_0)=c_n n!

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