Product rule for higher derivatives

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This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules

Statement

This states that if f and g are n times differentiable functions at x=x0, then the pointwise product f⋅g is also n times differentiable at x=x0, and we have:

dndxn[f(x)g(x)]|x=x0=∑k=0n(nk)f(k)(x0)g(n−k)(x0)

Here, f(k) denotes the kth derivative of f, g(n−k) denotes the (n−k)th derivative of g, and (nk) is the binomial coefficient. These are the same as the coefficients that appear in the expansion of (A+B)n.

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

dndxn[f(x)g(x)]=∑k=0n(nk)f(k)(x)g(n−k)(x)

Particular cases

Value of n Formula for dndxn[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)+2f′(x)g′(x)+f(x)g″(x).
3 f‴(x)g(x)+3f″(x)g′(x)+3f′(x)g″(x)+g‴(x).