# How do you find the n-th derivative of a power series?

Sep 12, 2014

If $f \left(x\right) = {\sum}_{k = 0}^{\infty} {c}_{k} {x}^{k}$, then
${f}^{\left(n\right)} \left(x\right) = {\sum}_{k = n}^{\infty} k \left(k - 1\right) \left(k - 2\right) \cdots \left(k - n + 1\right) {c}_{k} {x}^{k - n}$

By taking the derivative term by term,
$f ' \left(x\right) = {\sum}_{k = 1}^{\infty} k {c}_{k} {x}^{k - 1}$
$f ' ' \left(x\right) = {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {c}_{k} {x}^{k - 2}$
$f ' ' ' \left(x\right) = {\sum}_{k = 3}^{\infty} k \left(k - 1\right) \left(k - 2\right) {c}_{k} {x}^{k - 3}$
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${f}^{\left(n\right)} \left(x\right) = {\sum}_{k = n}^{\infty} k \left(k - 1\right) \left(k - 2\right) \cdots \left(k - n + 1\right) {c}_{k} {x}^{k - n}$