Product rule for higher derivatives - Revision history

Vipul: /* Statement */

2024-04-11T06:09:01Z

Statement

← Older revision		Revision as of 06:09, 11 April 2024
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	\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>		\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>
	\|-		\|-
	\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math>		\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>(fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math>
	\|-		\|-
	\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math>		\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math>

Vipul: /* Statement */

2024-04-11T06:08:21Z

Statement

← Older revision		Revision as of 06:08, 11 April 2024
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	! Version type !! Statement		! Version type !! Statement
	\|-		\|-
	\| specific point, named functions \|\| This states that if <math>f</math> and <math>g</math> are <math>n</math> times differentiable functions at <math>x = x_0</math>, then the [[pointwise product of functions\|pointwise product]] <math>f \cdot g</math> is also <math>n</math> times differentiable at <math>x = x_0</math>, and we have:<br><math>\frac{d^n}{dx^n}[f(x)g(x)]\|_{x = x_0} = \sum_{k=0}^n \binom{n}{k} f^{(n - k)}(x_0)g^{(k)}(x_0)</math><br>Here, <math>f^{(n - k)}</math> denotes the <math>(n-k)^{th}</math> derivative of <math>f</math> (with <math>f^{(0)} = f, f^{(1)} = f'</math>, etc.), <math>g^{(k)}</math> denotes the <math>k^{th}</math> derivative of <math>g</math>, and <math>\binom{n}{k}</math> is the [[binomial coefficient]]. These are the same as the coefficients that appear in the expansion of <math>\! (A + B)^n</math>.		\| specific point, named functions \|\| This states that if <math>f</math> and <math>g</math> are <math>n</math> times differentiable functions at <math>x = x_0</math>, then the [[pointwise product of functions\|pointwise product]] <math>f \cdot g</math> is also <math>n</math> times differentiable at <math>x = x_0</math>, and we have:<br><math>\frac{d^n}{dx^n}[f(x)g(x)]\|_{x = x_0} = \sum_{k=0}^n \binom{n}{k} f^{(n - k)}(x_0)g^{(k)}(x_0)</math><br>Here, <math>f^{(n - k)}</math> denotes the <math>(n-k)^{th}</math> derivative of <math>f</math> (with <math>f^{(0)} = f, f^{(1)} = f'</math>, etc.), <math>g^{(k)}</math> denotes the <math>k^{th}</math> derivative of <math>g</math>, and <math>\binom{n}{k}</math> is the [[binomial coefficient]]. These are the same as the coefficients that appear in the expansion of <math>(A + B)^n</math>.
	\|-		\|-
	\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>		\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>

Vipul at 05:59, 11 April 2024

2024-04-11T05:59:45Z

← Older revision		Revision as of 05:59, 11 April 2024
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			{{perspectives}}
	{{differentiation rule}}		{{differentiation rule}}

Vipul: /* Statement */

2012-06-30T15:38:11Z

Statement

← Older revision		Revision as of 15:38, 30 June 2012
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	\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>		\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>
	\|-		\|-
	\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! (~~f \cdot g~~)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math>		\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math>
	\|-		\|-
	\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math>		\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math>

Vipul: /* Statement */

2012-03-22T14:21:46Z

Statement

← Older revision		Revision as of 14:21, 22 March 2012
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	! Version type !! Statement		! Version type !! Statement
	\|-		\|-
	\| specific point, named functions \|\| This states that if <math>f</math> and <math>g</math> are <math>n</math> times differentiable functions at <math>x = x_0</math>, then the [[pointwise product of functions\|pointwise product]] <math>f \cdot g</math> is also <math>n</math> times differentiable at <math>x = x_0</math>, and we have:<br><math>\frac{d^n}{dx^n}[f(x)g(x)]\|_{x = x_0} = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x_0)g^{(n-k)}(x_0)</math><br>Here, <math>f^{(k)}</math> denotes the <math>k^{th}</math> derivative of <math>f</math> (with <math>f^{(0)} = f, f^{(1)} = f'</math>, etc.), <math>g^{(n-k)}</math> denotes the <math>~~(n-~~k)^{th}</math> derivative of <math>g</math>, and <math>\binom{n}{k}</math> is the [[binomial coefficient]]. These are the same as the coefficients that appear in the expansion of <math>\! (A + B)^n</math>.		\| specific point, named functions \|\| This states that if <math>f</math> and <math>g</math> are <math>n</math> times differentiable functions at <math>x = x_0</math>, then the [[pointwise product of functions\|pointwise product]] <math>f \cdot g</math> is also <math>n</math> times differentiable at <math>x = x_0</math>, and we have:<br><math>\frac{d^n}{dx^n}[f(x)g(x)]\|_{x = x_0} = \sum_{k=0}^n \binom{n}{k} f^{(n - k)}(x_0)g^{(k)}(x_0)</math><br>Here, <math>f^{(n - k)}</math> denotes the <math>(n-k)^{th}</math> derivative of <math>f</math> (with <math>f^{(0)} = f, f^{(1)} = f'</math>, etc.), <math>g^{(k)}</math> denotes the <math>k^{th}</math> derivative of <math>g</math>, and <math>\binom{n}{k}</math> is the [[binomial coefficient]]. These are the same as the coefficients that appear in the expansion of <math>\! (A + B)^n</math>.
	\|-		\|-
	\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x)g^{(n-k)}(x)</math>		\| generic point, named functions, point notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! \frac{d^n}{dx^n}[f(x)g(x)] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)</math>
	\|-		\|-
	\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! (f \cdot g)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(k)}g^{(n-k)}</math>		\| generic point, named functions, point-free notation \|\| If <math>f</math> and <math>g</math> are functions of one variable, the following holds wherever the right side makes sense:<br><math>\! (f \cdot g)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math>
	\|-		\|-
	\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^ku}{(dx)^k}\frac{d^~~{n-k}v~~}{(dx)^~~{n-~~k}}</math>		\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}u}{(dx)^{n-k}}\frac{d^kv}{(dx)^k}</math>
	\|}		\|}

Vipul: /* Particular cases */

2011-12-05T17:03:07Z

Particular cases

@@ Line 34: / Line 34: @@
 | 5 || <math>\! f^{(5)}(x)g(x) + 5f^{(4)}(x) g'(x) + 10f'''(x)g''(x) + 10f''(x)g'''(x) + 5f(x)g^{(4)}(x) + g^{(5)}(x)</math>
 |}
 ==Significance==

Vipul: /* Computational feasibility significance */

2011-11-27T21:58:04Z

Computational feasibility significance

← Older revision		Revision as of 21:58, 27 November 2011
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	===Computational feasibility significance===		===Computational feasibility significance===

	Each of the ~~version~~ has its own computational feasibility significance~~===~~		Each of the versions has its own computational feasibility significance:

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

Vipul: /* Particular cases */

2011-10-15T16:56:07Z

Particular cases

@@ Line 33: / Line 33: @@
 |-
 | 5 || <math>\! f^{(5)}(x)g(x) + 5f^{(4)}(x) g'(x) + 10f'''(x)g''(x) + 10f''(x)g'''(x) + 5f(x)g^{(4)}(x) + g^{(5)}(x)</math>
 |}

Vipul: /* Statement */

2011-10-15T16:42:48Z

Statement

← Older revision		Revision as of 16:42, 15 October 2011
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	\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^ku}{(dx)^k}\frac{d^{n-k}v}{(dx)^{n-k}}</math>		\| Pure Leibniz notation \|\| Suppose <math>u</math> and <math>v</math> are both variables functionally dependent on <math>x</math>. Then <math>\frac{d^n(uv)}{(dx)^n} = \sum_{k=0}^n \binom{n}{k} \frac{d^ku}{(dx)^k}\frac{d^{n-k}v}{(dx)^{n-k}}</math>
	\|}		\|}

			===One-sided version===

			There are analogues of each of the statements with one-sided derivatives. {{fillin}}

	==Particular cases==		==Particular cases==

Vipul: /* Particular cases */

2011-10-15T16:39:31Z

Particular cases

← Older revision		Revision as of 16:39, 15 October 2011
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	\|-		\|-
	\| 3 \|\| <math>\! f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + g'''(x)</math>.		\| 3 \|\| <math>\! f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + g'''(x)</math>.
			\|-
			\| 4 \|\| <math>\! f^{(4)}(x)g(x) + 4f'''(x)g'(x) + 6f''(x)g''(x) + 4f'(x)g'''(x) + f(x)g^{(4)}(x)</math>
			\|-
			\| 5 \|\| <math>\! f^{(5)}(x)g(x) + 5f^{(4)}(x) g'(x) + 10f'''(x)g''(x) + 10f''(x)g'''(x) + 5f(x)g^{(4)}(x) + g^{(5)}(x)</math>
	\|}		\|}