Question

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

Answer 1

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.