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 fg 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(nk)(x0)

Here, f(k) denotes the kth derivative of f, g(nk) denotes the (nk)th derivative of g, and (nk) is the binomial coefficient.

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(nk)(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).