Product rule for higher derivatives: Difference between revisions

Latest revision as of 06:09, 11 April 2024

ORIGINAL FULL PAGE: Product rule for higher derivatives
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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

Statement

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

One-sided version

There are analogues of each of the statements with one-sided derivatives. Fill this in later

Particular cases

Value of $n$	Formula for ${\frac {d^{n}}{dx^{n}}}[f(x)g(x)]$
1	$\!f'(x)g(x)+f(x)g'(x)$ (this is the usual product rule for differentiation).
2	$\!f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$ .
3	$\!f'''(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x)+g'''(x)$ .
4	$\!f^{(4)}(x)g(x)+4f'''(x)g'(x)+6f''(x)g''(x)+4f'(x)g'''(x)+f(x)g^{(4)}(x)$
5	$\!f^{(5)}(x)g(x)+5f^{(4)}(x)g'(x)+10f'''(x)g''(x)+10f''(x)g'''(x)+5f(x)g^{(4)}(x)+g^{(5)}(x)$

Related rules

Similar rules in single variable calculus

Similar rules in multivariable calculus

Product rule for higher partial derivatives

Similar rules in advanced mathematics

Binomial formula for powers of a derivation

Significance

Qualitative and existential significance

Each of the versions has its own qualitative significance:

Version type	Significance
specific point, named functions	This tells us that if $f$ and $g$ are both $n$ times differentiable at a point $x_{0}$ , so is $f\cdot g$ .
generic point, named functions, point notation	This tells us that if $f$ and $g$ are both $n$ times differentiable on an open interval, so is $f\cdot g$ .
generic point, named functions, point-free notation	This shows that the way that $(f\cdot g)^{(n)}$ behaves is governed by the nature of the derivatives (up to the $n^{th}$ ) of $f$ and $g$ . In particular, if $f^{(n)}$ and $g^{(n)}$ are both continuous functions on an interval, so is $(f\cdot g)^{(n)}$ .

Computational feasibility significance

Each of the versions has its own computational feasibility significance:

Version type	Significance
specific point, named functions	This tells us that knowing the values (in the sense of numerical values) of $f,f',f'',\dots ,f^{(n)}$ and $g,g',g'',\dots ,g^{(n)}$ at a point $x_{0}$ allows us to compute the value $(f\cdot g)^{(n)}(x_{0})$ by plugging into the formula and doing a bunch of multiplications and additions.
generic point, named functions	This tells us that knowledge of the generic expressions for $f,f',f'',\dots ,f^{(n)}$ and $g,g',g'',\dots ,g^{(n)}$ allows us to compute the generic expression for $(f\cdot g)^{(n)}$ .

@@ Line 1: / Line 1: @@
+{{perspectives}}
 {{differentiation rule}}
 ==Statement==
-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:
+{| class="sortable" border="1"
+! Version type !! Statement
-<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>
+|-
+| 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>.
-Here, <math>f^{(k)}</math> denotes the <math>k^{th}</math> derivative of <math>f</math>, <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>.
+|-
+| 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>
+|-
+| 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>
+|}
-If we consider this as a general expression rather than evaluating at a given point, we get:
+===One-sided version===
-<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>
+There are analogues of each of the statements with one-sided derivatives. {{fillin}}
 ==Particular cases==
@@ Line 23: / Line 30: @@
 |-
 | 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>
+|}
+==Related rules==
+===Similar rules in single variable calculus===
+* [[Product rule for differentiation]]
+* [[Chain rule for higher derivatives]]
+* [[Chain rule for differentiation]]
+* [[Repeated differentiation is linear]]
+===Similar rules in multivariable calculus===
+* [[Product rule for higher partial derivatives]]
+===Similar rules in advanced mathematics===
+* [[Groupprops:Binomial formula for powers of a derivation|Binomial formula for powers of a derivation]]
+==Significance==
+===Qualitative and existential significance===
+Each of the versions has its own qualitative significance:
+{| class="sortable" border="1"
+! Version type !! Significance
+|-
+| specific point, named functions || This tells us that if <math>f</math> and <math>g</math> are both <math>n</math> times differentiable at a point <math>x_0</math>, so is <math>f \cdot g</math>.
+|-
+| generic point, named functions, point notation || This tells us that if <math>f</math> and <math>g</math> are both <math>n</math> times differentiable on an [[open interval]], so is <math>f \cdot g</math>.
+|-
+| generic point, named functions, point-free notation || This shows that the way that <math>(f \cdot g)^{(n)}</math> behaves is governed by the nature of the derivatives (up to the <math>n^{th}</math>) of <math>f</math> and <math>g</math>. In particular, if <math>f^{(n)}</math> and <math>g^{(n)}</math> are both continuous functions on an interval, so is <math>(f \cdot g)^{(n)}</math>.
+|}
+===Computational feasibility significance===
+Each of the versions has its own computational feasibility significance:
+{| class="sortable" border="1"
+! Version type !! Significance
+|-
+| specific point, named functions || This tells us that knowing the values (in the sense of ''numerical values'') of <math>f,f',f'',\dots, f^{(n)}</math> and <math>g,g',g'',\dots,g^{(n)}</math> at a point <math>x_0</math> allows us to compute the value <math>(f \cdot g)^{(n)}(x_0)</math> by plugging into the formula and doing a bunch of multiplications and additions.
+|-
+| generic point, named functions || This tells us that knowledge of the ''generic'' expressions for <math>f,f',f'',\dots, f^{(n)}</math> and <math>g,g',g'',\dots,g^{(n)}</math> allows us to compute the generic expression for <math>(f \cdot g)^{(n)}</math>.
 |}