Differentiation is linear: Difference between revisions

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{{differentiation rule}}
{{differentiation rule}}
{{linear operator statement for functions}}


==Statement==
==Statement==
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<math>\frac{d}{dx}[f(x) + g(x)]_{x = x_0} = f'(x_0) + g'(x_0)</math>
<math>\frac{d}{dx}[f(x) + g(x)]_{x = x_0} = f'(x_0) + g'(x_0)</math>


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


<math>(f + g)'(x_0) = f'(x_0) + g'(x_0)</math>
<math>(f + g)'(x_0) = f'(x_0) + g'(x_0)</math>


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


<math>(f + g)' = f' + g'</math>
<math>(f + g)' = f' + g'</math>
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===In terms of generalized linearity===
===In terms of generalized linearity===


{{fillin}}
Suppose <math>f_1, f_2, \dots, f_n</math> are functions that are all differentiable at a point <math>x_0</math> and <math>a_1, a_2, \dots, a_n</math> are real numbers. Then:
 
<math>\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)</math>


==Related rules==
==Related rules==
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* [[Definite integration is linear]]
* [[Definite integration is linear]]
* [[Indefinite integration is linear]]
* [[Indefinite integration is linear]]
==Facts used==
# [[uses::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''': <math>f</math> and <math>g</math> are functions that are both differentiable at <math>x = x_0</math>.
'''To prove''': <math>f + g</math> is differentiable at <math>x = x_0</math>, and <math>\! (f + g)'(x_0) = f'(x_0) + g'(x_0)</math>
'''Proof''': Our proof strategy is to start out by trying to compute <math>\! (f + g)'(x_0)</math> as a difference quotient, and keep simplifying this, using Fact (1) in the process.
{{fillin}}

Latest revision as of 14:53, 24 September 2021

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.

Fill this in later