Differentiation is linear

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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:

$(f + g)'(x_0) = f'(x_0) + g'(x_0)$

In point-free notation:

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

• 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:

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

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).

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.