Repeated differentiation is linear

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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:

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