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

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Answer 1 · 2014-09-27T12:26:36

One of the benefits of a Taylor series is the ease of differentiation since it can be done term by term. So, the Talyor series

$f(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}(x-a)^n$

can be differentiated as

$f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(x-a)^{n}]'$

by Power Rule,

$=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(x-a)^{n-1}]$

(Note: $n$ starts from $1$ since the term is zero when $n=0$ )
by simplifying a little further,

$=sum_{n=1}^infty{f^{(n)}(a)}/{(n-1)!}(x-a)^{n-1}$

I hope that this was helpful.