Product rule for higher derivatives: Difference between revisions

Revision as of 16:37, 15 October 2011

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.
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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: $\frac{d^{n}}{d x^{n}} [f (x) g (x)] \|_{x = x_{0}} = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x_{0}) g^{(n - k)} (x_{0})$ Here, $f^{(k)}$ denotes the $k^{t h}$ derivative of $f$ (with $f^{(0)} = f, f^{(1)} = f^{'}$ , etc.), $g^{(n - k)}$ denotes the $(n - k)^{t h}$ 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: $\frac{d^{n}}{d x^{n}} [f (x) g (x)] = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x) g^{(n - k)} (x)$
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: $(f \cdot g)^{(n)} = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} g^{(n - k)}$
Pure Leibniz notation	Suppose $u$ and $v$ are both variables functionally dependent on $x$ . Then $\frac{d^{n} (u v)}{(d x)^{n}} = \sum_{k = 0}^{n} (\binom{n}{k}) \frac{d^{k} u}{(d x)^{k}} \frac{d^{n - k} v}{(d x)^{n - k}}$

Particular cases

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

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 ==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^{(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>.
-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^{(k)}(x)g^{(n-k)}(x)</math>
-If we consider this as a general expression rather than evaluating at a given point, we get:
+|-
+| 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>
-<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>
+|-
+| 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>
+|}
 ==Particular cases==