Product rule for differentiation - Revision history

Vipul: /* Sanity checks */

2021-09-21T01:29:00Z

Sanity checks

← Older revision		Revision as of 01:29, 21 September 2021
Line 247:		Line 247:
	\| <math>g = 1/f</math> \|\| zero function \|\| The product is <math>1</math>, which is a constant function, so its derivative is zero. \|\| We get <math>f(x)(1/f)'(x) + f'(x)/f(x)</math>. By the chain rule, <math>(1/f)'(x) = -f'(x)/(f(x))^2</math>, so plugging in, we get <math>-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)</math>, which simplifies to zero.		\| <math>g = 1/f</math> \|\| zero function \|\| The product is <math>1</math>, which is a constant function, so its derivative is zero. \|\| We get <math>f(x)(1/f)'(x) + f'(x)/f(x)</math>. By the chain rule, <math>(1/f)'(x) = -f'(x)/(f(x))^2</math>, so plugging in, we get <math>-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)</math>, which simplifies to zero.
	\|}		\|}

	~~<center>{{#widget:YouTube\|id=pS3iUxkReAI}}</center>~~

	<section end="mathmajor"/>		<section end="mathmajor"/>

Vipul: /* Compatibility checks */

2021-09-21T01:26:45Z

Compatibility checks

← Older revision		Revision as of 01:26, 21 September 2021
Line 165:		Line 165:

	<math>((f_1 \cdot f_2) \cdot f_3)' = (f_1 \cdot f_2)' \cdot f_3 + (f_1 \cdot f_2) \cdot f_3' = (f_1' \cdot f_2 + f_1 \cdot f_2') \cdot f_3 + f_1 \cdot f_2 \cdot f_3' = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'</math>		<math>((f_1 \cdot f_2) \cdot f_3)' = (f_1 \cdot f_2)' \cdot f_3 + (f_1 \cdot f_2) \cdot f_3' = (f_1' \cdot f_2 + f_1 \cdot f_2') \cdot f_3 + f_1 \cdot f_2 \cdot f_3' = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'</math>

	~~<center>{{#widget:YouTube\|id=WFufWneBoec}}</center>~~

	===Compatibility with linearity===		===Compatibility with linearity===

Vipul: /* Qualitative and existential significance */

2021-09-21T01:25:44Z

Qualitative and existential significance

← Older revision		Revision as of 01:25, 21 September 2021
Line 112:		Line 112:
	\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].		\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].
	\|}		\|}

	~~<center>{{#widget:YouTube\|id=WFufWneBoec}}</center>~~

	<section begin="school"/>		<section begin="school"/>

Vipul: /* Statement in multiple versions */

2021-09-21T01:25:29Z

Statement in multiple versions

← Older revision		Revision as of 01:25, 21 September 2021
Line 31:		Line 31:
	\|}		\|}

	~~<center>{{#widget:YouTube\|id=hmoNCJxyb4QY}}</center>~~
	{{generic point specific point confusion}}		{{generic point specific point confusion}}

Vipul: /* Statement for multiple functions */

2021-09-21T01:25:11Z

Statement for multiple functions

← Older revision		Revision as of 01:25, 21 September 2021
Line 67:		Line 67:

	<math>\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)</math>		<math>\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)</math>

	~~<center>{{#widget:YouTube\|id=-ghadmPQxvI}}</center>~~

	<section end="college"/>		<section end="college"/>

Vipul: /* Qualitative and existential significance */

2014-10-05T15:26:52Z

Qualitative and existential significance

← Older revision		Revision as of 15:26, 5 October 2014
Line 113:		Line 113:
	\| generic point, named functions, point notation \|\| This tells us that if both <math>f</math> and <math>g</math> are differentiable on an open interval, then so is <math>f \cdot g</math>. The one-sided versions allow us to make similar statements for closed intervals where we require the appropriate one-sided differentiability at the endpoints.		\| generic point, named functions, point notation \|\| This tells us that if both <math>f</math> and <math>g</math> are differentiable on an open interval, then so is <math>f \cdot g</math>. The one-sided versions allow us to make similar statements for closed intervals where we require the appropriate one-sided differentiability at the endpoints.
	\|-		\|-
	\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)'</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].		\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].
	\|}		\|}

Vipul: /* Statement for two functions */

2012-04-09T03:47:48Z

Statement for two functions

← Older revision		Revision as of 03:47, 9 April 2012
Line 9:		Line 9:
	<section begin="school"/>		<section begin="school"/>
	==Statement for two functions==		==Statement for two functions==

	~~<center>{{#widget:YouTube\|id=85jXtIRg1ro}}</center>~~

	===Statement in multiple versions===		===Statement in multiple versions===

Vipul at 03:40, 9 April 2012

2012-04-09T03:40:07Z

← Older revision		Revision as of 03:40, 9 April 2012
Line 9:		Line 9:
	<section begin="school"/>		<section begin="school"/>
	==Statement for two functions==		==Statement for two functions==

			<center>{{#widget:YouTube\|id=85jXtIRg1ro}}</center>

	===Statement in multiple versions===		===Statement in multiple versions===
Line 31:		Line 33:
	\|}		\|}

			<center>{{#widget:YouTube\|id=hmoNCJxyb4QY}}</center>
	{{generic point specific point confusion}}		{{generic point specific point confusion}}

Line 66:		Line 69:

	<math>\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)</math>		<math>\! (f_1 \cdot f_2 \cdot f_3)'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)</math>

			<center>{{#widget:YouTube\|id=-ghadmPQxvI}}</center>

	<section end="college"/>		<section end="college"/>
Line 112:		Line 117:
	\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)'</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].		\| generic point, point-free notation \|\| This can be used to deduce more, namely that the nature of <math>(f \cdot g)'</math> depends strongly on the nature of <math>f</math> and that of <math>g</math>. In particular, if <math>f</math> and <math>g</math> are both [[continuously differentiable function]]s on an interval (i.e., <math>f'</math> and <math>g'</math> are both continuous on that interval), then <math>(f \cdot g)'</math> is also continuously differentiable on that interval. This uses the [[sum theorem for continuity]] and [[product theorem for continuity]].
	\|}		\|}

			<center>{{#widget:YouTube\|id=WFufWneBoec}}</center>

	<section begin="school"/>		<section begin="school"/>
Line 165:		Line 172:

	<math>((f_1 \cdot f_2) \cdot f_3)' = (f_1 \cdot f_2)' \cdot f_3 + (f_1 \cdot f_2) \cdot f_3' = (f_1' \cdot f_2 + f_1 \cdot f_2') \cdot f_3 + f_1 \cdot f_2 \cdot f_3' = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'</math>		<math>((f_1 \cdot f_2) \cdot f_3)' = (f_1 \cdot f_2)' \cdot f_3 + (f_1 \cdot f_2) \cdot f_3' = (f_1' \cdot f_2 + f_1 \cdot f_2') \cdot f_3 + f_1 \cdot f_2 \cdot f_3' = f_1' \cdot f_2 \cdot f_3 + f_1 \cdot f_2' \cdot f_3 + f_1 \cdot f_2 \cdot f_3'</math>

			<center>{{#widget:YouTube\|id=WFufWneBoec}}</center>

	===Compatibility with linearity===		===Compatibility with linearity===
Line 247:		Line 256:
	\| <math>g = 1/f</math> \|\| zero function \|\| The product is <math>1</math>, which is a constant function, so its derivative is zero. \|\| We get <math>f(x)(1/f)'(x) + f'(x)/f(x)</math>. By the chain rule, <math>(1/f)'(x) = -f'(x)/(f(x))^2</math>, so plugging in, we get <math>-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)</math>, which simplifies to zero.		\| <math>g = 1/f</math> \|\| zero function \|\| The product is <math>1</math>, which is a constant function, so its derivative is zero. \|\| We get <math>f(x)(1/f)'(x) + f'(x)/f(x)</math>. By the chain rule, <math>(1/f)'(x) = -f'(x)/(f(x))^2</math>, so plugging in, we get <math>-f(x)f'(x)/(f(x))^2 + f'(x)/f(x)</math>, which simplifies to zero.
	\|}		\|}

			<center>{{#widget:YouTube\|id=pS3iUxkReAI}}</center>

	<section end="mathmajor"/>		<section end="mathmajor"/>

Vipul: /* Sanity checks */

2012-04-02T22:25:33Z

Sanity checks

← Older revision		Revision as of 22:25, 2 April 2012
Line 234:		Line 234:
	===Sanity checks===		===Sanity checks===

	We first consider examples where the product rule for differentiation confirms something we already knew through other means:		We first consider examples where the product rule for differentiation confirms something we already knew through other means. In all examples, we assume that both <math>f</math> and <math>g</math> are differentiable functions:

	{\| class="sortable" border="1"		{\| class="sortable" border="1"

Vipul: /* Case of infinite or undefined values */

2012-04-02T04:51:11Z

Case of infinite or undefined values

← Older revision		Revision as of 04:51, 2 April 2012
Line 203:		Line 203:
	==Case of infinite or undefined values==		==Case of infinite or undefined values==

	{{further\|[[Using the product rule for differentiation for limiting behavior ~~in cases of~~ undefined derivative]]}}		{{further\|[[Using the product rule for differentiation for limiting behavior at points with undefined derivative]]}}

	The product rule for differentiation has analogues for infinities, with the appropriate caveats about indeterminate forms. Specifically, we have the following:		The product rule for differentiation has analogues for infinities, with the appropriate caveats about indeterminate forms. Specifically, we have the following: