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{{about||Euler's chain rule relating partial derivatives of three independent variables|Triple product rule|the counting principle in combinatorics|Rule of product|the general product rule in probability|Chain rule (probability)}}
{{short description|Formula for the derivative of a product}}
{{Calculus |Differential}}

In [[calculus]], the '''product rule''' is a formula used to find the [[derivative]]s of products of two or more [[Functions (mathematics)|functions]]. It may be stated as

:<math>(f\cdot g)'=f'\cdot g+f\cdot g' </math>

or in [[Leibniz's notation]]

:<math>\dfrac{d}{dx}(u\cdot v) = \dfrac{du}{dx} \cdot v + u\cdot \dfrac{dv}{dx}.</math>

In differentials notation, this can be written as
:<math> d(uv)=u\,dv+v\,du.</math>

In Leibniz's notation, the derivative of the product of three functions (not to be confused with [[Triple product rule|Euler's triple product rule]]) is

:<math>\dfrac{d}{dx}(u\cdot v \cdot w)=\dfrac{du}{dx} \cdot v \cdot w + u \cdot \dfrac{dv}{dx} \cdot w + u\cdot v\cdot \dfrac{dw}{dx}.</math>

== Discovery ==
Discovery of this rule is credited to [[Gottfried Leibniz]], who demonstrated it using [[differential (calculus)|differentials]].<ref>{{cite journal |author=Michelle Cirillo
|journal=The Mathematics Teacher |volume=101 |issue=1 |pages=23&ndash;27 |date=August 2007 |title=Humanizing Calculus
|url=http://www.nctm.org/publications/article.aspx?id=19302
|url-access=subscription }}</ref>  (However, J. M. Child, a translator of Leibniz's papers,<ref>{{citation|first=G. W.|last=Leibniz|url=http://dynref.engr.illinois.edu/rvc_Child_1920.pdf|title= The Early Mathematical Manuscripts of Leibniz|translator=J.M. Child|year=2005|origyear=1920|publisher=Dover|page= 28, footnote 58|isbn=978-0-486-44596-0}}</ref> argues that it is due to [[Isaac Barrow]].) Here is Leibniz's argument: Let ''u''(''x'') and ''v''(''x'') be two [[differentiable function]]s of ''x''. Then the differential of ''uv'' is

: <math>
\begin{align}
d(u\cdot v) & {} = (u + du)\cdot (v + dv) - u\cdot v \\
& {} = u\cdot dv + v\cdot du + du\cdot dv.
\end{align}
</math>

Since the term ''du''·''dv'' is "negligible" (compared to ''du'' and ''dv''), Leibniz concluded that

:<math>d(u\cdot v) = v\cdot du + u\cdot dv </math>

and this is indeed the differential form of the product rule. If we divide through by the differential ''dx'', we obtain

:<math>\frac{d}{dx} (u\cdot v) = v \cdot \frac{du}{dx} + u \cdot  \frac{dv}{dx} </math>

which can also be written in [[Derivative#Lagrange.27s notation|Lagrange's notation]] as

:<math>(u\cdot v)' = v\cdot  u' + u\cdot  v'. </math>

== Examples ==
*Suppose we want to differentiate ''f''(''x'') = ''x''<sup>2</sup> sin(''x''). By using the product rule, one gets the derivative ''{{prime|f}}''(''x'') = 2''x'' sin(''x'') + ''x''<sup>2</sup> cos(''x'') (since the derivative of ''x''<sup>2</sup> is 2''x'' and the derivative of the [[sine]] function is the cosine function).
*One special case of the product rule is the [[constant multiple rule]], which states: if ''c'' is a number and ''f''(''x'') is a differentiable function, then ''cf''(''x'') is also differentiable, and its derivative is (''cf''{{prime|)}}(''x'') = ''c{{prime|f}}''(''x''). This follows from the product rule since the derivative of any constant is zero. This, combined with the sum rule for derivatives, shows that differentiation is [[linear transformation|linear]].
*The rule for [[integration by parts]] is derived from the product rule, as is (a weak version of) the [[quotient rule]]. (It is a "weak" version in that it does not prove that the quotient is differentiable, but only says what its derivative is {{em|if}} it is differentiable.)

== Proofs ==

=== Proof by factoring (from first principles) ===

Let {{math|''h''(''x'') {{=}} ''f''(''x'')''g''(''x'')}} and suppose that {{mvar|f}} and {{mvar|g}} are each differentiable at {{mvar|x}}. We want to prove that {{mvar|h}} is differentiable at {{mvar|x}} and that its derivative, {{math|''{{prime|h}}''(''x'')}}, is given by {{math|''{{prime|f}}''(''x'')''g''(''x'') + ''f''(''x'')''{{prime|g}}''(''x'')}}. To do this, <math>f(x)g(x+\Delta x)-f(x)g(x+\Delta x)</math> (which is zero, and thus does not change the value) is added to the numerator to permit its factoring, and then properties of limits are used.

:<math>\begin{align}
 h'(x) &= \lim_{\Delta x\to 0} \frac{h(x+\Delta x)-h(x)}{\Delta x} \\[5pt]
   &= \lim_{\Delta x\to 0} \frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x} \\[5pt]
   &= \lim_{\Delta x\to 0} \frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x} \\[5pt]
   &= \lim_{\Delta x\to 0} \frac{\big[f(x+\Delta x)-f(x)\big] \cdot g(x+\Delta x) + f(x) \cdot \big[g(x+\Delta x)-g(x)\big]}{\Delta x} \\[5pt]
   &= \lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \cdot \underbrace{\lim_{\Delta x\to 0} g(x+\Delta x)}_\text{See the note below.}
     + \lim_{\Delta x\to 0} f(x) \cdot \lim_{\Delta x\to 0} \frac{g(x+\Delta x)-g(x)}{\Delta x} \\[5pt]
   &= f'(x)g(x)+f(x)g'(x).
\end{align}</math>
The fact that
: <math> \lim_{\Delta x\to0} g(x+\Delta x) = g(x) </math>
is deduced from a theorem that states that differentiable functions are continuous.

=== Brief proof ===
By definition, if <math> f, g: \mathbb{R} \rightarrow \mathbb{R} </math> are differentiable at <math> x </math> then we can write
:<math> f(x+h) = f(x) + f'(x)h + \psi_1(h) \qquad \qquad g(x+h) = g(x) + g'(x)h + \psi_2(h) </math>
such that <math> \lim_{h \to 0} \frac{\psi_1(h)}{h} = \lim_{h \to 0} \frac{\psi_2(h)}{h} = 0, </math> [[Big O notation#Little-o notation|also written]] <math>\psi_1, \psi_2 \sim o(h)</math>. Then:

:<math> \begin{align} fg(x+h) - fg(x) =  (f(x) + f'(x)h +\psi_1(h))(g(x) + g'(x)h + \psi_2(h)) - fg(x)= f'(x)g(x)h + f(x)g'(x)h + o(h) \\[12pt] \end{align} </math>
Taking the limit for small <math> h </math> gives the result.

=== Quarter squares ===
There is a proof using [[Multiplication algorithm#Quarter square multiplication|quarter square multiplication]] which relies on the [[chain rule]] and on the properties of the quarter square function (shown here as ''q'', i.e., with <math>q(x)=\tfrac{x^2}4</math>):

:<math>f=q(u+v)-q(u-v), </math>

Differentiating both sides:

:<math>\begin{align}
f' &= q'(u+v)(u'+v') - q'(u-v)(u'-v') \\[4pt]
&= \left({1 \over 2}(u+v)(u'+v')\right) - \left({1 \over 2}(u-v)(u'-v')\right) \\[4pt]
&= {1 \over 2}(uu' + vu' + uv' + vv') - {1 \over 2}(uu' - vu' - uv' + vv') \\[4pt]
&= vu'+uv' \\[4pt]
&= uv'+u'v
\end{align}</math>

=== Chain rule ===
The product rule can be considered a special case of the [[chain rule]] for several variables.

: <math> {d (ab) \over dx} = \frac{\partial(ab)}{\partial a}\frac{da}{dx}+\frac{\partial (ab)}{\partial b}\frac{db}{dx} = b \frac{da}{dx} + a \frac{db}{dx}. </math>

=== Non-standard analysis ===
Let ''u'' and ''v'' be continuous functions in ''x'', and let ''dx'', ''du'' and ''dv'' be [[infinitesimal]]s within the framework of [[non-standard analysis]], specifically the [[hyperreal number]]s. Using st to denote the [[standard part function]] that associates to a [[wiktionary:Finite number|finite]] hyperreal number the real infinitely close to it, this gives

:<math>\begin{align}
   \frac{d(uv)}{dx} &= \operatorname{st}\left(\frac{(u + du)(v + dv) - uv}{dx}\right) \\[4pt]
   &= \operatorname{st}\left(\frac{uv + u \cdot dv + v \cdot du + dv \cdot du -uv}{dx}\right) \\[4pt]
   &= \operatorname{st}\left(\frac{u \cdot dv + (v + dv) \cdot du}{dx}\right) \\[4pt]
   &= u \frac{dv}{dx} + v \frac{du}{dx}.
 \end{align}</math>

This was essentially [[Leibniz]]'s proof exploiting the [[transcendental law of homogeneity]] (in place of the standard part above).

===Smooth infinitesimal analysis===
In the context of Lawvere's approach to infinitesimals, let ''dx'' be a nilsquare infinitesimal. Then ''du''&nbsp;=&nbsp;''u''&prime;&nbsp;''dx'' and ''dv''&nbsp;=&nbsp;''v''&prime;&nbsp;''dx'', so that

: <math> 
\begin{align}
d(uv) & = (u + du)(v + dv)  -uv \\
 & = uv + u\cdot dv + v\cdot du + du\cdot dv - uv \\
 & = u\cdot dv + v\cdot du + du\cdot dv \\
 & = u\cdot dv + v\cdot du\,\!
\end{align}
</math>

since

:<math>du \, dv = u' v' (dx)^2 = 0</math>

== Generalizations ==
{{unreferenced section|date=July 2013}}

=== A product of more than two factors ===
The product rule can be generalized to products of more than two factors. For example, for three factors we have

:<math>\frac{d(uvw)}{dx} = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}.</math>

For a collection of functions <math>f_1, \dots, f_k</math>, we have

:<math>\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ]
 = \sum_{i=1}^k \left(\left(\frac{d}{dx} f_i(x) \right) \prod_{j\ne i} f_j(x) \right)
= \left(  \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right).</math>

=== Higher derivatives ===
{{Main|General Leibniz rule}}
It can also be generalized to the [[general Leibniz rule]] for the ''n''th derivative of a product of two factors, by symbolically expanding according to the [[binomial theorem]]:
:<math>d^n(uv) = \sum_{k=0}^n {n \choose k} \cdot d^{(n-k)}(u)\cdot  d^{(k)}(v).</math>

Applied at a specific point ''x'', the above formula gives: 
:<math>(uv)^{(n)}(x) = \sum_{k=0}^n {n \choose k} \cdot u^{(n-k)}(x)\cdot  v^{(k)}(x).</math>

Furthermore, for the ''n''th derivative of an arbitrary number of factors:
:<math>\left(\prod_{i=1}^kf_i\right)^{(n)}=\sum_{j_1+j_2+\cdots+j_k=n}{n\choose j_1,j_2,\ldots,j_k}\prod_{i=1}^kf_i^{(j_i)}.</math>

=== Higher partial derivatives ===
For [[partial derivative]]s, we have

:<math>{\partial^n \over \partial x_1\,\cdots\,\partial x_n} (uv)
= \sum_S {\partial^{|S|} u \over \prod_{i\in S} \partial x_i} \cdot {\partial^{n-|S|} v \over \prod_{i\not\in S} \partial x_i}</math>

where the index {{mvar|S}} runs through all {{math|2<sup>''n''</sup>}} [[subset]]s of {{math|{{mset|1, ..., ''n''}}}}, and {{math|{{mabs|''S''}}}} is the [[cardinality]] of {{mvar|S}}. For example, when {{math|1=''n'' = 3}},

:<math>\begin{align} & {\partial^3 \over \partial x_1\,\partial x_2\,\partial x_3} (uv)  \\[6pt]
= {} & u \cdot{\partial^3 v \over \partial x_1\,\partial x_2\,\partial x_3} + {\partial u \over \partial x_1}\cdot{\partial^2 v \over \partial x_2\,\partial x_3} +  {\partial u \over \partial x_2}\cdot{\partial^2 v \over \partial x_1\,\partial x_3} + {\partial u \over \partial x_3}\cdot{\partial^2 v \over \partial x_1\,\partial x_2} \\[6pt]
&  + {\partial^2 u \over \partial x_1\,\partial x_2}\cdot{\partial v \over \partial x_3}
+ {\partial^2 u \over \partial x_1\,\partial x_3}\cdot{\partial v \over \partial x_2}
+ {\partial^2 u \over \partial x_2\,\partial x_3}\cdot{\partial v \over \partial x_1}
+ {\partial^3 u \over \partial x_1\,\partial x_2\,\partial x_3}\cdot v. \end{align}</math>

=== Banach space ===
Suppose ''X'', ''Y'', and ''Z'' are [[Banach space]]s (which includes [[Euclidean space]]) and ''B'' : ''X'' &times; ''Y'' → ''Z'' is a [[continuous function (topology)|continuous]] [[bilinear operator]]. Then ''B'' is differentiable, and its derivative at the point (''x'',''y'') in ''X'' &times; ''Y'' is the [[linear map]] ''D''<sub>(''x'',''y'')</sub>''B'' : ''X'' &times; ''Y'' → ''Z'' given by
:<math> (D_\left( x,y \right)\,B)\left( u,v \right) = B\left( u,y \right) + B\left( x,v \right)\qquad\forall (u,v)\in X \times Y. </math>

=== Derivations in abstract algebra ===
In [[abstract algebra]], the product rule is used to ''define'' what is called a [[derivation (abstract algebra)|derivation]], not vice versa.

=== Vector functions ===
The product rule extends to [[scalar multiplication]], [[dot product]]s, and [[cross product]]s of vector functions.

For scalar multiplication:
<math>(f \cdot \mathbf g)' = f\;'\cdot \mathbf g + f \cdot \mathbf g\;' </math>

For dot products:
<math>(\mathbf f \cdot \mathbf g)' = \mathbf f\;'\cdot \mathbf g + \mathbf f \cdot \mathbf g\;' </math>

For cross products:
<math>(\mathbf f \times \mathbf g)' = \mathbf f\;' \times \mathbf g + \mathbf f \times \mathbf g\;' </math>

Note: cross products are not commutative, i.e. <math>(f \times g)' \neq f' \times g + g' \times f </math>, instead products are anticommutative, so it can be written as <math>(f \times g)' = f' \times g - g' \times f </math>

=== Scalar fields ===
For scalar fields the concept of [[gradient]] is the analog of the derivative:

<math>\nabla (f \cdot g) = \nabla f \cdot g + f \cdot \nabla g </math>

== Applications ==
Among the applications of the product rule is a proof that

:<math> {d \over dx} x^n = nx^{n-1}</math>

when ''n'' is a positive integer (this rule is true even if ''n'' is not positive or is not an integer, but the proof of that must rely on other methods). The proof is by [[mathematical induction]] on the exponent ''n''. If ''n''&nbsp;=&nbsp;0 then ''x''<sup>''n''</sup> is constant and ''nx''<sup>''n''&nbsp;&minus;&nbsp;1</sup> = 0. The rule holds in that case because the derivative of a constant function is&nbsp;0. If the rule holds for any particular exponent ''n'', then for the next value, ''n''&nbsp;+&nbsp;1, we have

:<math>\begin{align}
{d \over dx}x^{n+1} &{}= {d \over dx}\left( x^n\cdot x\right) \\[12pt]
&{}= x{d \over dx} x^n + x^n{d \over dx}x \qquad\mbox{(the product rule is used here)} \\[12pt]
&{}= x\left(nx^{n-1}\right) + x^n\cdot 1\qquad\mbox{(the induction hypothesis is used here)} \\[12pt]
&{}= (n + 1)x^n.
\end{align} </math>

Therefore, if the proposition is true for ''n'', it is true also for&nbsp;''n''&nbsp;+&nbsp;1, and therefore for all natural ''n''.

==References==
{{reflist}}


{{use dmy dates|date=January 2012}}

[[Category:Differentiation rules]]
[[Category:Articles containing proofs]]