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{{short description|Formula for the derivative of a product}}
[[File:Schema_Règle_produit.png|thumb|300px|right|Geometric illustration of a proof of the product rule]]
{{Calculus |Differential}}

In [[Calculus|calculus]], the '''product rule''' (or '''Leibniz rule'''<ref>{{Cite web|url=https://encyclopediaofmath.org/wiki/Leibniz_rule|title=Leibniz rule – Encyclopedia of Mathematics}}</ref> or '''Leibniz product rule''') is a formula used to find the [[Derivative|derivative]]s of products of two or more functions. For two functions, it may be stated in [[Notation for differentiation#Lagrange's notation|Lagrange's notation]] as <math display="block">(u \cdot v)' = u ' \cdot v + u \cdot v' </math> or in [[Leibniz's notation]] as <math display="block">\frac{d}{dx} (u\cdot v) = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}.</math>

The rule may be extended or generalized to products of three or more functions, to a rule for higher-order derivatives of a product, and to other contexts.

== Discovery ==
Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials.<ref>{{cite journal |author=Michelle Cirillo
|journal=The Mathematics Teacher |volume=101 |issue=1 |pages=23–27 |date=August 2007 |title=Humanizing Calculus
|doi=10.5951/MT.101.1.0023 |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|orig-year=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|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 <math>f(x)=x^2\text {sin}(x).</math> By using the product rule, one gets the derivative <math>f'(x)=2x\cdot \text {sin}(x)+x^2\text {cos} (x)</math> (since the derivative of <math>x^2</math> is <math>2x,</math> and the derivative of the [[Sine|sine]] function is the cosine function).
*One special case of the product rule is the [[Constant multiple rule|constant multiple rule]], which states: if {{mvar|c}} is a number, and <math>f(x)</math> is a differentiable function, then <math>c\cdot f(x)</math> is also differentiable, and its derivative is <math>(cf)'(x)=c \cdot f'(x).</math> 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.
*The rule for [[Integration by parts|integration by parts]] is derived from the product rule, as is (a weak version of) the [[Quotient rule|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 ==

=== Limit definition of derivative ===

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> follows from the fact that differentiable functions are continuous.
=== Linear approximations ===
By definition, if <math> f, g: \mathbb{R} \to \mathbb{R} </math> are differentiable at <math> x </math>, then we can write [[Linear approximation|linear approximations]]:
<math display="block"> f(x+h) = f(x) + f'(x)h + \varepsilon_1(h) </math> and <math display="block"> g(x+h) = g(x) + g'(x)h + \varepsilon_2(h), </math>
where the error terms are small with respect to ''h'': that is, <math display="inline"> \lim_{h \to 0} \frac{\varepsilon_1(h)}{h} = \lim_{h \to 0} \frac{\varepsilon_2(h)}{h} = 0, </math> [[Big O notation#Little-o notation|also written]] <math>\varepsilon_1, \varepsilon_2 \sim o(h)</math>. Then:
<math display="block"> \begin{align} 
   f(x+h)g(x+h) - f(x)g(x) &=  (f(x) + f'(x)h +\varepsilon_1(h))(g(x) + g'(x)h + \varepsilon_2(h)) - f(x)g(x) \\[.5em]
                   &= f(x)g(x) + f'(x)g(x)h + f(x)g'(x)h -f(x)g(x) + \text{error terms} \\[.5em]
                   &= f'(x)g(x)h + f(x)g'(x)h + o(h) . 
\end{align} </math>
The "error terms" consist of items such as <math>f(x)\varepsilon_2(h), f'(x)g'(x)h^2</math> and <math>hf'(x)\varepsilon_1(h)</math> which are easily seen to have magnitude <math>o(h).</math> Dividing by <math> h </math> and taking the limit <math> h\to 0 </math> gives the result.

=== Quarter squares ===
This proof uses the [[Chain rule|chain rule]] and the [[Multiplication algorithm#Quarter square multiplication|quarter square function]] <math>q(x)=\tfrac14x^2</math> with derivative <math>q'(x) = \tfrac12 x</math>. We have:
:<math>uv=q(u+v)-q(u-v), </math>

and differentiating both sides gives:

:<math>\begin{align}
f' &= q'(u+v)(u'+v') - q'(u-v)(u'-v') \\[4pt]
&= \left(\tfrac12(u+v)(u'+v')\right) - \left(\tfrac12(u-v)(u'-v')\right) \\[4pt]
&= \tfrac12(uu' + vu' + uv' + vv') - \tfrac12(uu' - vu' - uv' + vv') \\[4pt]
&= vu'+uv' .
\end{align}</math>

=== Multivariable chain rule ===
The product rule can be considered a special case of the [[Chain rule|chain rule]] for several variables, applied to the multiplication function <math>m(u,v)=uv</math>:

: <math> {d (uv) \over dx} = \frac{\partial(uv)}{\partial u}\frac{du}{dx}+\frac{\partial (uv)}{\partial v}\frac{dv}{dx} = v \frac{du}{dx} + u \frac{dv}{dx}. </math>

=== Non-standard analysis ===
Let ''u'' and ''v'' be continuous functions in ''x'', and let ''dx'', ''du'' and ''dv'' be [[Infinitesimal|infinitesimal]]s within the framework of [[Non-standard analysis|non-standard analysis]], specifically the [[Hyperreal number|hyperreal number]]s. Using st to denote the [[Standard part function|standard part function]] that associates to a 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) \\
   &= \operatorname{st}\left(\frac{uv + u \cdot dv + v \cdot du + du \cdot dv -uv}{dx}\right) \\
   &= \operatorname{st}\left(\frac{u \cdot dv + v \cdot du + du \cdot dv}{dx}\right) \\
   &= \operatorname{st}\left(u \frac{dv}{dx} + (v + dv) \frac{du}{dx}\right) \\
   &= 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 <math>dx</math> be a nilsquare infinitesimal. Then <math>du = u'\ dx</math> and <math>dv = v'\ dx</math>, 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> Dividing by <math>dx</math> then gives <math>\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}</math> or <math>(uv)' = u \cdot v' + v \cdot u'</math>.

===Logarithmic differentiation===

Let <math>h(x) = f(x) g(x)</math>. Taking the [[Absolute value|absolute value]] of each function and the [[Natural logarithm|natural log]] of both sides of the equation, 

: <math>\ln|h(x)| = \ln|f(x) g(x)| </math>

Applying properties of the absolute value and logarithms, 

: <math>\ln|h(x)| = \ln|f(x)| + \ln|g(x)| </math> 

Taking the [[Logarithmic derivative|logarithmic derivative]] of both sides and then solving for <math>h'(x) </math>:

: <math>
\frac{h'(x)}{h(x)} = \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}
</math>

Solving for <math>h'(x)</math> and substituting back <math>f(x) g(x)</math> for <math>h(x)</math> gives:

: <math>
\begin{align}
h'(x) &= h(x)\left(\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}\right) \\
&= f(x) g(x)\left(\frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)}\right) \\
&= f'(x) g(x) + f(x) g'(x).
\end{align}
</math>

Note: Taking the absolute value of the functions is necessary for the [[Logarithmic differentiation|logarithmic differentiation]] of functions that may have negative values, as logarithms are only [[Real-valued function|real-valued]] for positive arguments. This works because <math>
\tfrac{d}{dx}(\ln |u|) = \tfrac{u'}{u}
</math>, which justifies taking the absolute value of the functions for logarithmic differentiation.

== Generalizations ==
=== 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=1,j\ne i}^k 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>

The [[Logarithmic derivative|logarithmic derivative]] provides a simpler expression of the last form, as well as a direct proof that does not involve any [[Recursion|recursion]]. The ''logarithmic derivative''   of a function {{mvar|f}}, denoted here {{math|Logder(''f'')}}, is the derivative of the [[Logarithm|logarithm]] of the function. It follows that 
:<math>\operatorname{Logder}(f)=\frac {f'}f.</math>
Using that the logarithm of a product is the sum of the logarithms of the factors, the [[Sum rule in differentiation|sum rule]] for derivatives gives immediately
:<math>\operatorname{Logder}(f_1\cdots f_k)= \sum_{i=1}^k\operatorname{Logder}(f_i).</math>
The last above expression of the derivative of a product is obtained by multiplying both members of this equation by the product of the <math>f_i.</math>

=== Higher derivatives ===
{{Main|General Leibniz rule}}
It can also be generalized to the [[General Leibniz rule|general Leibniz rule]] for the ''n''th derivative of a product of two factors, by symbolically expanding according to the [[Binomial theorem|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, one has a similar formula with [[Multinomial theorem|multinomial coefficients]]:
:<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|partial derivative]]s, we have<ref>{{cite journal |author=Micheal Hardy
|journal=The Electronic Journal of Combinatorics |volume=13 |date=January 2006 |title=Combinatorics of Partial Derivatives
|arxiv=math/0601149 |bibcode=2006math......1149H |url=https://www.emis.de//journals/EJC/Volume_13/PDF/v13i1r1.pdf
}}</ref>

:<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|subset]]s of {{math|{{mset|1, ..., ''n''}}}}, and {{math|{{mabs|''S''}}}} is the [[Cardinality|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 bilinear operator. Then ''B'' is differentiable, and its derivative at the point (''x'',''y'') in ''X'' &times; ''Y'' is the [[Linear map|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>

This result can be extended<ref>{{cite book |last1=Kreigl |first1=Andreas |last2=Michor |first2=Peter |title=The Convenient Setting of Global Analysis |date=1997 |publisher=American Mathematical Society |isbn=0-8218-0780-3 |page=59 |url=https://www.mat.univie.ac.at/~michor/apbookh-ams.pdf}}</ref> to more general topological vector spaces.

=== In vector calculus ===

The product rule extends to various product operations of vector functions on <math>\mathbb{R}^n</math>:<ref>{{citation|last = Stewart | first = James|title = Calculus | edition = 8 | publisher = Cengage|year = 2016}}, Section 13.2.</ref>

* For [[Scalar multiplication|scalar multiplication]]: <math display="block">(f \cdot \mathbf g)' = f'\cdot \mathbf g + f \cdot \mathbf g' </math>
* For [[Dot product|dot product]]: <math display="block">(\mathbf f \cdot \mathbf g)' = \mathbf f' \cdot \mathbf g + \mathbf f \cdot \mathbf g' </math>
* For [[Cross product|cross product]] of vector functions on <math>\mathbb{R}^3</math>: <math display="block">(\mathbf f \times \mathbf g)' = \mathbf f' \times \mathbf g + \mathbf f \times \mathbf g' </math>

There are also analogues for other analogs of the derivative: if ''f'' and ''g'' are scalar fields then there is a product rule with the [[Gradient|gradient]]:

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

Such a rule will hold for any continuous [[Bilinear map|bilinear]] product operation.  Let ''B'' : ''X'' &times; ''Y'' → ''Z'' be a continuous bilinear map between vector spaces, and let ''f'' and ''g'' be differentiable functions into ''X'' and ''Y'', respectively.  The only properties of multiplication used in the proof using the limit definition of derivative is that multiplication is continuous and bilinear.  So for any continuous bilinear operation,
<math display="block">H(f, g)' = H(f', g) + H(f, g').</math>
This is also a special case of the product rule for bilinear maps in Banach space.

=== Derivations in abstract algebra and differential geometry ===
In [[Abstract algebra|abstract algebra]], the product rule is the defining property of a derivation. In this terminology, the product rule states that the derivative operator is a derivation  on functions. 

In [[Differential geometry|differential geometry]], a [[Tangent vector|tangent vector]] to a [[Manifold|manifold]] ''M'' at a point ''p'' may be defined abstractly as an operator on real-valued functions which behaves like a [[Directional derivative|directional derivative]] at ''p'': that is, a [[Linear form|linear functional]] ''v'' which is a derivation,<blockquote><math>v(fg) = v(f)\,g(p) + f(p) \, v(g).</math></blockquote>
Generalizing (and dualizing) the formulas of vector calculus to an ''n''-dimensional manifold ''M,'' one may take [[Differential form|differential forms]] of degrees ''k'' and ''l'', denoted <math>\alpha\in \Omega^k(M), \beta\in \Omega^\ell(M)</math>, with the wedge or [[Exterior algebra|exterior product]] operation <math>\alpha\wedge\beta\in \Omega^{k+\ell}(M)</math>, as well as the [[Exterior derivative|exterior derivative]] <math>d:\Omega^m(M)\to\Omega^{m+1}(M)</math>. Then one has the [[Differential graded algebra|graded Leibniz rule]]:<blockquote><math>d(\alpha\wedge\beta)= d\alpha \wedge \beta + (-1)^{k} \alpha\wedge d\beta.</math></blockquote>

== 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|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''.

==See also==

* {{annotated link|Differentiation of integrals}}
* {{annotated link|Differentiation of trigonometric functions}}
* {{annotated link|Differentiation rules}}
* {{annotated link|Distribution (mathematics)}}
* {{annotated link|General Leibniz rule}}
* {{annotated link|Integration by parts}}
* {{annotated link|Inverse functions and differentiation}}
* {{annotated link|Linearity of differentiation}}
* {{annotated link|Power rule}}
* {{annotated link|Quotient rule}}
* {{annotated link|Vector calculus identities}}

==References==
{{reflist}}

{{Calculus topics}}

[[Category:Articles containing proofs]]
[[Category:Differentiation rules]]
[[Category:Theorems in analysis]]
[[Category:Theorems in calculus]]

{{Sourceattribution|Product rule}}