Repeated differentiation is linear

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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.
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For k a positive integer, denote by f^{(k)} the function obtained by differentiating f a total of k times. The operation f \mapsto f^{(k)} is a linear operator. We give two equivalent ways of stating this below.

In terms of additivity and pulling out scalars

The following are true:

  • Repeated differentiation is additive, or k^{th} derivative of sum is sum of derivatives: If f and g are functions that are both differentiable at x = x_0, we have:

\frac{d^k}{dx^k}[f(x) + g(x)]_{x = x_0} = f^{(k)}(x_0) + g^{(k)}(x_0)

or equivalently:

(f + g)^{(k)}(x_0) = f^{(k)}(x_0) + g^{(k)}(x_0)

In point-free notation:

(f + g)^{(k)} = f^{(k)} + g^{(k)}

  • Constants (also called scalars) can be pulled out of differentiations: If f is differentiable at x = x_0 and \lambda is a real number, then:

\frac{d^k}{dx^k}[\lambda f(x)]|_{x = x_0} = \lambda f^{(k)}(x_0)

In terms of generalized linearity

Suppose f_1, f_2, \dots, f_n are functions that are all differentiable at a point x_0 and a_1, a_2, \dots, a_n are real numbers. Then:

\frac{d^k}{dx^k}[a_1f_1(x) + a_2f_2(x) + \dots + a_nf_n(x)]|_{x = x_0} = a_1f_1^{(k)}(x_0) + a_2f_2^{(k)}(x_0) + \dots + a_nf_n^{(k)}(x_0)

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