Product rule for differentiation: Difference between revisions

Revision as of 11:31, 26 August 2011

Statement for two functions

Verbal statement

If two (possibly equal) functions are differentiable at a given real number, then their pointwise product is also differentiable at that number and the derivative of the product is the sum of two terms: the derivative of the first function times the second function and the first function times the derivative of the second function.

Statement with symbols

Suppose $f$ and $g$ are functions, both of which are differentiable at a real number $x = x_{0}$ . Then, the product function $f \cdot g$ , defined as $x \mapsto f (x) g (x)$ is also differentiable at $x$ , and the derivative at $x_{0}$ is given as follows:

$\frac{d}{d x} [f (x) g (x)] |_{x = x_{0}} = f^{'} (x_{0}) g (x_{0}) + f (x_{0}) g^{'} (x_{0})$

or equivalently:

$\frac{d}{d x} [f (x) g (x)] |_{x = x_{0}} = \frac{d (f (x))}{d x} |_{x = x_{0}} \cdot g (x_{0}) + f (x_{0}) \cdot \frac{d (g (x))}{d x} |_{x = x_{0}}$

If we consider the general expressions rather than evaluation at a particular point $x_{0}$ , we can rewrite the above as:

$\frac{d}{d x} [f (x) g (x)] = f^{'} (x) g (x) + f (x) g^{'} (x)$

or equivalently:

$(f \cdot g)^{'} = (f^{'} \cdot g) + (f \cdot g^{'})$

Statement for multiple functions

If $f_{1}, f_{2}, \dots, f_{n}$ are all functions, and we define $F (x) : = f_{1} (x) f_{2} (x) \dots f_{n} (x)$ , then we have:

$F^{'} (x) = {f_{1^{'}}}^{'} (x) f_{2} (x) \dots f_{n} (x) + f_{1} (x) {f_{2^{'}}}^{'} (x) \dots f_{n} (x) + \dots + f_{1} (x) f_{2} (x) \dots f_{n - 1} (x) {f_{n^{'}}}^{'} (x)$

In other words, we get a sum of $n$ terms, each of which is a product of $n$ evaluations, of which only one is a derivative, and the one we choose as the derivative cycles through all the $n$ possibilities.

For instance, if $n = 3$ , we get:

$F^{'} (x) = {f_{1^{'}}}^{'} (x) f_{2} (x) f_{3} (x) + f_{1} (x) {f_{2^{'}}}^{'} (x) f_{3} (x) + f_{1} (x) f_{2} (x) {f_{3^{'}}}^{'} (x)$

Examples

Trivial examples

We first consider examples where the product rule for differentiation confirms something we already knew through other means:

Case	What we know about the derivative of $x \mapsto f (x) g (x)$	What we know about $f^{'} (x) g (x) + f (x) g^{'} (x)$
$g$ is the zero function.	The derivative is the zero function, because $f (x) g (x) = 0$ for all $x$ .	Both $g (x)$ and $g^{'} (x)$ are zero functions, so $f^{'} (x) g (x) + f (x) g^{'} (x)$ is everywhere zero.
$g$ is a constant nonzero function with value $λ$ .	The derivative is $λ f^{'} (x)$ , because the constant can be pulled out of the differentiation process.	$f^{'} (x) g (x)$ simplifies to $λ f^{'} (x)$ . Since $g$ is constant, $g^{'} (x)$ is the zero function, hence so is $f (x) g^{'} (x)$ . The sum is thus $λ f^{'} (x)$ .
$f = g$	The derivative is $2 f (x) f^{'} (x)$ by the chain rule for differentiation: we are composing the square function and $f$ .	We get $f^{'} (x) f (x) + f (x) f^{'} (x) = 2 f (x) f^{'} (x)$ .

Nontrivial examples where simple alternate methods exist

Fill this in later

Nontrivial examples where simple alternate methods do not exist

Fill this in later

@@ Line 22: / Line 22: @@
 <math>(f \cdot g)' = (f' \cdot g) + (f \cdot g')</math>
+==Statement for multiple functions==
+If <math>f_1,f_2,\dots,f_n</math> are all functions, and we define <math>F(x) := f_1(x)f_2(x) \dots f_n(x)</math>, then we have:
+<math>F'(x) = f_1'(x)f_2(x) \dots f_n(x) + f_1(x)f_2'(x) \dots f_n(x) + \dots + f_1(x)f_2(x) \dots f_{n-1}(x)f_n'(x)</math>
+In other words, we get a sum of <math>n</math> terms, each of which is a product of <math>n</math> evaluations, of which only ''one'' is a derivative, and the one we choose as the derivative cycles through all the <math>n</math> possibilities.
+For instance, if <math>n = 3</math>, we get:
+<math>\! F'(x) = f_1'(x)f_2(x)f_3(x) + f_1(x)f_2'(x)f_3(x) + f_1(x)f_2(x)f_3'(x)</math>
+==Examples==
+===Trivial examples===
+We first consider examples where the product rule for differentiation confirms something we already knew through other means:
+{| class="sortable" border="1"
+! Case !! What we know about the derivative of <math>x \mapsto f(x)g(x)</math> !! What we know about <math>f'(x)g(x) + f(x)g'(x)</math>
+|-
+| <math>g</math> is the [[zero function]]. || The derivative is the zero function, because <math>f(x)g(x) = 0</math> for all <math>x</math>. || Both <math>g(x)</math> and <math>g'(x)</math> are zero functions, so <math>f'(x)g(x) + f(x)g'(x)</math> is everywhere zero.
+|-
+| <math>g</math> is a constant nonzero function with value <math>\lambda</math>. || The derivative is <math>\lambda f'(x)</math>, because the constant can be ''pulled'' out of the differentiation process. || <math>f'(x)g(x)</math> simplifies to <math>\lambda f'(x)</math>. Since <math>g</math> is constant, <math>g'(x)</math> is the zero function, hence so is <math>f(x)g'(x)</math>. The sum is thus <math>\lambda f'(x)</math>.
+|-
+| <math>f = g</math> || The derivative is <math>2f(x)f'(x)</math> by the [[chain rule for differentiation]]: we are composing the [[square function]] and <math>f</math>. || We get <math>f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x)</math>.
+|}
+===Nontrivial examples where simple alternate methods exist===
+{{fillin}}
+===Nontrivial examples where simple alternate methods do not exist===
+{{fillin}}