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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This article gives a statement of the form that a certain operator from a space of functions to another space of functions is a linear operator, i.e., applying the operator to the sum of two functions gives the sum of the applications to each function, and applying it to a scalar multiple of a function gives the same scalar multiple of its application to the function.

Statement

In terms of additivity and pulling out scalars

The following are true:

Differentiation is additive, or derivative of sum is sum of derivatives: If $f$ and $g$ are functions that are both differentiable at $x=x_{0}$ , we have:

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

or equivalently, the following holds whenever the right side is defined (see concept of equality conditional to existence of one side):

$(f+g)'(x_{0})=f'(x_{0})+g'(x_{0})$

In point-free notation, we have the following whereever the right side is defined (see concept of equality conditional to existence of one side):

$(f+g)'=f'+g'$

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:

${\frac {d}{dx}}[\lambda f(x)]|_{x=x_{0}}=\lambda f'(x_{0})$

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:

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

Related rules

Facts used

Limit is linear: This states that the limit of the sum is the sum of the limits and scalars can be pulled out of limits.

Proof

We prove here the two-sided versions. Analogous proofs exist for the one-sided versions, and these use the one-sided versions of Fact (1).

Proof of additivity

Given: $f$ and $g$ are functions that are both differentiable at $x=x_{0}$ .

To prove: $f+g$ is differentiable at $x=x_{0}$ , and $\!(f+g)'(x_{0})=f'(x_{0})+g'(x_{0})$

Proof: Our proof strategy is to start out by trying to compute $\!(f+g)'(x_{0})$ as a difference quotient, and keep simplifying this, using Fact (1) in the process.

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