Differentiation is linear - Revision history

Vipul: /* In terms of additivity and pulling out scalars */

2021-09-24T14:53:13Z

In terms of additivity and pulling out scalars

← Older revision		Revision as of 14:53, 24 September 2021
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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>

Vipul: /* Proof of additivity */

2011-10-20T21:48:46Z

Proof of additivity

← Older revision		Revision as of 21:48, 20 October 2011
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	'''Given''': <math>f</math> and <math>g</math> are functions that are both differentiable at <math>x = x_0</math>.		'''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>		'''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.		'''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}}		{{fillin}}

Vipul at 21:47, 20 October 2011

2011-10-20T21:47:34Z

← Older revision		Revision as of 21:47, 20 October 2011
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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}}

Vipul at 02:03, 7 September 2011

2011-09-07T02:03:08Z

← Older revision		Revision as of 02:03, 7 September 2011
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	{{differentiation rule}}		{{differentiation rule}}

			{{linear operator statement for functions}}

	==Statement==		==Statement==

Vipul: /* Statement */

2011-09-05T13:58:53Z

Statement

← Older revision		Revision as of 13:58, 5 September 2011
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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==

Vipul: /* Related rules */

2011-09-05T13:57:11Z

Related rules

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	==Related rules==		==Related rules==

			* [[Repeated differentiation is linear]]
	* [[Product rule for differentiation]]		* [[Product rule for differentiation]]
	* [[Product rule for higher derivatives]]		* [[Product rule for higher derivatives]]

Vipul at 13:54, 5 September 2011

2011-09-05T13:54:36Z

← Older revision		Revision as of 13:54, 5 September 2011
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			{{differentiation rule}}

	==Statement==		==Statement==

Vipul: Created page with "==Statement== ===In terms of additivity and pulling out scalars=== The following are true: * ''Differentiation is additive'', or ''derivative of sum is sum of derivatives'': I..."

2011-08-26T13:33:36Z

Created page with "==Statement== ===In terms of additivity and pulling out scalars=== The following are true: * ''Differentiation is additive'', or ''derivative of sum is sum of derivatives'': I..."

New page

==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 <math>f</math> and <math>g</math> are functions that are both differentiable at <math>x = x_0</math>, we have:

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

or equivalently:

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

In point-free notation:

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

* ''Constants (also called scalars) can be pulled out of differentiations'': If <math>f</math> is differentiable at <math>x = x_0</math> and <math>\lambda</math> is a real number, then:

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

===In terms of generalized linearity===

{{fillin}}

==Related rules==

* [[Product rule for differentiation]]
* [[Product rule for higher derivatives]]
* [[Chain rule for differentiation]]
* [[Chain rule for higher derivatives]]
* [[Definite integration is linear]]
* [[Indefinite integration is linear]]