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 $k^{th}$ 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:

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

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