Product rule for higher derivatives: Difference between revisions

Latest revision as of 06:09, 11 April 2024

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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 $(f\cdot g)^{(n)}$ .

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 ! 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-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>