Repeated differentiation is linear

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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Statement

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 $k^{t h}$ derivative of sum is sum of derivatives: If $f$ and $g$ are functions that are both differentiable at $x = x_{0}$ , we have:

$\frac{d^{k}}{d x^{k}} [f (x) + g (x)]_{x = x_{0}} = f^{(k)} (x_{0}) + g^{(k)} (x_{0})$

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 $λ$ is a real number, then:

$\frac{d^{k}}{d x^{k}} [λ f (x)] |_{x = x_{0}} = λ f^{(k)} (x_{0})$

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:

$\frac{d^{k}}{d x^{k}} [a_{1} f_{1} (x) + a_{2} f_{2} (x) + \dots + a_{n} f_{n} (x)] |_{x = x_{0}} = a_{1} f_{1}^{(k)} (x_{0}) + a_{2} f_{2}^{(k)} (x_{0}) + \dots + a_{n} f_{n}^{(k)} (x_{0})$

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