Differentiation is linear

From Calculus

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.
View other differentiation rules

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=x0, we have:

ddx[f(x)+g(x)]x=x0=f(x0)+g(x0)

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

(f+g)(x0)=f(x0)+g(x0)

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=x0 and λ is a real number, then:

ddx[λf(x)]|x=x0=λf(x0)

In terms of generalized linearity

Suppose f1,f2,,fn are functions that are all differentiable at a point x0 and a1,a2,,an are real numbers. Then:

ddx[a1f1(x)+a2f2(x)++anfn(x)]|x=x0=a1f1(x0)+a2f2(x0)++anfn(x0)

Related rules

Facts used

  1. 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=x0.

To prove: f+g is differentiable at x=x0, and (f+g)(x0)=f(x0)+g(x0)

Proof: Our proof strategy is to start out by trying to compute (f+g)(x0) as a difference quotient, and keep simplifying this, using Fact (1) in the process.

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