Product rule for higher derivatives

Statement

This states that if $f$ and $g$ are $n$ times differentiable functions at $x = x_{0}$ , then the pointwise product $f \cdot g$ is also $n$ times differentiable at $x = x_{0}$ , and we have:

$\frac{d^{n}}{d x^{n}} [f (x) g (x)] |_{x = x_{0}} = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x_{0}) g^{(n - k)} (x_{0})$

Here, $f^{(k)}$ denotes the $k^{t h}$ derivative of $f$ , $g^{(n - k)}$ denotes the $(n - k)^{t h}$ derivative of $g$ , and $(\binom{n}{k})$ is the binomial coefficient.

If we consider this as a general expression rather than evaluating at a given point, we get:

$\frac{d^{n}}{d x^{n}} [f (x) g (x)] = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x) g^{(n - k)} (x)$

Particular cases

Value of $n$	Formula for $\frac{d^{n}}{d x^{n}} [f (x) g (x)]$
1	$f^{'} (x) g (x) + f (x) g^{'} (x)$ (this is the usual product rule for differentiation).
2	$f^{″} (x) g (x) + 2 f^{'} (x) g^{'} (x) + f (x) g^{″} (x)$ .
3	$f^{‴} (x) g (x) + 3 f^{″} (x) g^{'} (x) + 3 f^{'} (x) g^{″} (x) + g^{‴} (x)$ .