Product rule for higher derivatives: Difference between revisions

Revision as of 16:31, 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

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 $f$ , $g^{(n - k)}$ 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 $(A + B)^{n}$ .

If we consider this as a general expression rather than evaluating at a given point, we get:

$\frac{d^{n}}{d x^{n}} [f (x) g (x)] = \sum_{k = 0}^{n} (\binom{n}{k}) f^{(k)} (x) g^{(n - k)} (x)$

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	$f^{‴} (x) g (x) + 3 f^{″} (x) g^{'} (x) + 3 f^{'} (x) g^{″} (x) + g^{‴} (x)$ .

@@ Line 7: / Line 7: @@
 <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>
-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]].
+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>.
 If we consider this as a general expression rather than evaluating at a given point, we get: