Chain rule for differentiation - Revision history

Vipul: /* Statement for two functions */

2023-12-05T07:06:42Z

Statement for two functions

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	\| generic point, named functions, point notation \|\| Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then, we have <br><math>\! \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)</math> wherever the right side expression makes sense.		\| generic point, named functions, point notation \|\| Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then, we have <br><math>\! \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)</math> wherever the right side expression makes sense.
	\|-		\|-
	\| generic point, named functions, point-free notation \|\| Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then,<br><math>\! (f \circ g)' = (f' \circ g) \cdot g'</math> ~~where the right side expression makes sense, where <math>\cdot</math> denotes the [[pointwise product of functions]].~~		\| generic point, named functions, point-free notation \|\| Suppose <math>f</math> and <math>g</math> are [[function]]s of one variable. Then, we have the following equality of functions on the omain where the right side expression makes sense, where <math>\cdot</math> denotes the [[pointwise product of functions]] (see [[concept of equality conditional to existence of one side]]):<br><math>\! (f \circ g)' = (f' \circ g) \cdot g'</math>
	\|-		\|-
	\| pure Leibniz notation \|\| Suppose <math>u = g(x)</math> is a function of <math>x</math> and <math>v = f(u)</math> is a function of <math>u</math>. Then, <br><math> \frac{dv}{dx} = \frac{dv}{du}\frac{du}{dx}</math>		\| pure Leibniz notation \|\| Suppose <math>u = g(x)</math> is a function of <math>x</math> and <math>v = f(u)</math> is a function of <math>u</math>. Then, <br><math> \frac{dv}{dx} = \frac{dv}{du}\frac{du}{dx}</math>

Vipul: /* Compatibility with notions of order */

2023-12-04T21:11:35Z

Compatibility with notions of order

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	* <math>(f \circ g)'</math> has order <math>mn -1</math>: First, note that <math>f \circ g</math> has order <math>mn</math> by the product relation for order. Next, note that differentiating pushes the order down by one.		* <math>(f \circ g)'</math> has order <math>mn -1</math>: First, note that <math>f \circ g</math> has order <math>mn</math> by the product relation for order. Next, note that differentiating pushes the order down by one.
	* <math>(f' \circ g) \cdot g'</math> has order <math>mn - 1</math>: Note that <math>f' \circ g</math> has order <math>(m - 1)n</math> and <math>g'</math> has order <math>n - 1</math>. Adding, we get <math(m - 1)n + n - 1 = mn - 1</math>.		* <math>(f' \circ g) \cdot g'</math> has order <math>mn - 1</math>: Note that <math>f' \circ g</math> has order <math>(m - 1)n</math> and <math>g'</math> has order <math>n - 1</math>. Adding, we get <math>(m - 1)n + n - 1 = mn - 1</math>.

	Note that this compatibility check ''fails'' on both the false chain rules discussed in [[#Significance\|the significance section]]: <toggledisplay>		Note that this compatibility check ''fails'' on both the false chain rules discussed in [[#Significance\|the significance section]]: <toggledisplay>

Vipul: /* Nontrivial examples */

2015-05-03T19:58:56Z

Nontrivial examples

← Older revision		Revision as of 19:58, 3 May 2015
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	===Nontrivial examples===		===Nontrivial examples===

	The chain rule is necessary for computing the derivatives of functions whose definition ''requires'' one to compose functions. The chain rule still isn't the only option: one can always compute the derivative as a limit of a [[difference quotient]]. But it does offer the only option if one restricts oneself to operating within the family of [[Category:Differentiation rules\|differentiation rules]].		The chain rule is necessary for computing the derivatives of functions whose definition ''requires'' one to compose functions. The chain rule still isn't the only option: one can always compute the derivative as a limit of a [[difference quotient]]. But it does offer the only option if one restricts oneself to operating within the family of [[:Category:Differentiation rules\|differentiation rules]].

	Some examples of functions for which the chain rule needs to be used include:		Some examples of functions for which the chain rule needs to be used include:

Vipul: /* Nontrivial examples */

2015-05-03T19:58:38Z

Nontrivial examples

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	===Nontrivial examples===		===Nontrivial examples===

	The chain rule is necessary for computing the derivatives of functions whose definition ''requires'' one to compose functions. The chain rule still isn't the only option: one can always compute the derivative as a limit of a [[difference quotient]]. But it does offer the only option if one restricts oneself to operating within the family of differentiation rules]].		The chain rule is necessary for computing the derivatives of functions whose definition ''requires'' one to compose functions. The chain rule still isn't the only option: one can always compute the derivative as a limit of a [[difference quotient]]. But it does offer the only option if one restricts oneself to operating within the family of [[Category:Differentiation rules\|differentiation rules]].

	Some examples of functions for which the chain rule needs to be used include:		Some examples of functions for which the chain rule needs to be used include:

Vipul: /* Computational feasibility significance */

2015-05-03T19:57:52Z

Computational feasibility significance

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	\| specific point, named functions \|\| If we know the values (''in the sense of numerical values'') <math>g'(x_0)</math> and <math>f'(g(x_0))</math>, we can use these to compute <math>(f \circ g)'(x_0)</math>.		\| specific point, named functions \|\| If we know the values (''in the sense of numerical values'') <math>g'(x_0)</math> and <math>f'(g(x_0))</math>, we can use these to compute <math>(f \circ g)'(x_0)</math>.
	\|-		\|-
	\| generic point, named functions \|\| This tells us that knowledge of the ''general expressions'' for the derivatives of <math>f</math> and <math>g</math> (along with expressions for the functions themselves) allows us to compute the general expression for the derivative of <math>f \circ g</math>.		\| generic point, named functions \|\| This tells us that knowledge of the ''general expressions'' for the derivatives of <math>f</math> and <math>g</math> (along with expressions for the functions themselves) allows us to compute the general expression for the derivative of <math>f \circ g</math>.<br>Note that we do not need to know <math>f</math> itself (it suffices to know <math>f'</math>, which tells us what <math>f</math> is up to additive constants), but we ''do'' need to know what <math>g</math> is. It does not suffice to know <math>g</math> merely up to additive constants.
	\|}		\|}

Vipul: /* Sine of sine function= */

2015-04-12T16:30:17Z

Sine of sine function=

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	{{#lst:sine of square function\|differentiation}}		{{#lst:sine of square function\|differentiation}}

	===Sine of sine function====		====Sine of sine function====

	Consider the [[sine of sine function]]:		Consider the [[sine of sine function]]:

Vipul: /* Sine of square function */

2015-04-12T16:30:02Z

Sine of square function

← Older revision		Revision as of 16:30, 12 April 2015
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	{{#lst:sine of square function\|differentiation}}		{{#lst:sine of square function\|differentiation}}

			===Sine of sine function====

			Consider the [[sine of sine function]]:

			<math>x \mapsto \sin(\sin x)</math>

			The derivative is:

			<math>(\sin \circ \sin)'(x) = (\sin' \circ \sin)(x)\sin'(x) = \cos(\sin x)\cos(x)</math>

Vipul: /* Nontrivial examples */

2015-04-12T16:27:25Z

Nontrivial examples

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	* An exponential function applied to a nonlinear algebraic function		* An exponential function applied to a nonlinear algebraic function
	* A composite of two trigonometric functions, two exponential functions, or an exponential and a trigonometric function		* A composite of two trigonometric functions, two exponential functions, or an exponential and a trigonometric function

			A few examples are below.

			====Sine of square function====

	Consider the [[sine of square function]]:		Consider the [[sine of square function]]:

Vipul: /* Nontrivial examples */

2015-04-12T16:27:07Z

Nontrivial examples

← Older revision		Revision as of 16:27, 12 April 2015
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	===Nontrivial examples===		===Nontrivial examples===

	~~Here are some~~ examples ~~that cannot be computed using methods other than~~ the chain rule:		The chain rule is necessary for computing the derivatives of functions whose definition ''requires'' one to compose functions. The chain rule still isn't the only option: one can always compute the derivative as a limit of a [[difference quotient]]. But it does offer the only option if one restricts oneself to operating within the family of differentiation rules]].

			Some examples of functions for which the chain rule needs to be used include:

			* A trigonometric function applied to a nonlinear algebraic function
			* An exponential function applied to a nonlinear algebraic function
			* A composite of two trigonometric functions, two exponential functions, or an exponential and a trigonometric function

	Consider the [[sine of square function]]:		Consider the [[sine of square function]]:

Vipul: /* Compatibility with notions of order */

2015-04-12T16:23:58Z

Compatibility with notions of order

← Older revision		Revision as of 16:23, 12 April 2015
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	* <math>(f \circ g)'</math> has order <math>mn -1</math>: First, note that <math>f \circ g</math> has order <math>mn</math> by the product relation for order. Next, note that differentiating pushes the order down by one.		* <math>(f \circ g)'</math> has order <math>mn -1</math>: First, note that <math>f \circ g</math> has order <math>mn</math> by the product relation for order. Next, note that differentiating pushes the order down by one.
	* <math>(f' \circ g) \cdot g'</math> has order <math>mn - 1</math>: Note that <math>f' \circ g</math> has order <math>(m - 1)n</math> and <math>g'</math> has order <math>n - 1</math>. Adding, we get <math(m - 1)n + n - 1 = mn - 1</math>.		* <math>(f' \circ g) \cdot g'</math> has order <math>mn - 1</math>: Note that <math>f' \circ g</math> has order <math>(m - 1)n</math> and <math>g'</math> has order <math>n - 1</math>. Adding, we get <math(m - 1)n + n - 1 = mn - 1</math>.

			Note that this compatibility check ''fails'' on both the false chain rules discussed in [[#Significance\|the significance section]]: <toggledisplay>

			* The rule <math>(f \circ g)'(x) = f'(g'(x))</math> fails because the order of the right side computes to <math>(m - 1)(n - 1)</math>, which is not the same as <math>mn - 1</math>.
			* The rule <math>(f \circ g)'(x) = f'(x)g'(x)</math> fails because the order of the right side computes to <math>m - 1 + n - 1 = m + n - 2</math>, which is not the same as <math>mn - 1</math>.</toggledisplay>

	Some examples of the notion of order which illustrate this are:		Some examples of the notion of order which illustrate this are: