Product rule

# Product rule

Discussion

Encyclopedia
{{dablink|For Euler's chain rule relating partial derivatives of three independent variables, see Triple product rule
Triple product rule
The triple product rule, known variously as the cyclic chain rule, cyclic relation, cyclical rule or Euler's chain rule, is a formula which relates partial derivatives of three interdependent variables...

.}} {{dablink|For the counting principle in combinatorics, see Rule of product
Rule of product
In combinatorics, the rule of product or multiplication principle is a basic counting principle...

.}} {{Calculus|cTopic=Differentiation}} In calculus
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related by the fundamental theorem...

, the product rule is a formula used to find the derivative
Derivative
In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a...

s of products of two or more functions. It may be stated thus: $\left(f\cdot g\right)\text{'}=f\text{'}\cdot g+f\cdot g\text{'} \,\!$ or in the Leibniz notation
Leibniz notation
In calculus, Leibniz's notation, named in honor of the 17th-century German philosopher and mathematician Gottfried Wilhelm Leibniz, uses the symbols dx and dy to represent "infinitely small" increments of x and y, just as Δx and Δy represent finite increments of x and y...

thus: $\dfrac\left\{d\right\}\left\{dx\right\}\left(u\cdot v\right)=u\cdot \dfrac\left\{dv\right\}\left\{dx\right\}+v\cdot \dfrac\left\{du\right\}\left\{dx\right\}$. The derivative of the product of three functions is: $\dfrac\left\{d\right\}\left\{dx\right\}\left(u\cdot v \cdot w\right)=\dfrac\left\{du\right\}\left\{dx\right\} \cdot v \cdot w + u \cdot \dfrac\left\{dv\right\}\left\{dx\right\} \cdot w + u\cdot v\cdot \dfrac\left\{dw\right\}\left\{dx\right\}$.

## Discovery by Leibniz

Discovery of this rule is credited to Gottfried Leibniz
Gottfried Leibniz
Gottfried Wilhelm Leibniz was a German philosopher and mathematician. He wrote in different languages, primarily in Latin , French and German ....

(however, Child (2008) argues that it is due to Isaac Barrow
Isaac Barrow
Isaac Barrow was an English Christian theologian, and mathematician who is generally given credit for his early role in the development of infinitesimal calculus; in particular, for the discovery of the fundamental theorem of calculus. His work centered on the properties of the tangent; Barrow was...

), who demonstrated it using differentials
Differential (calculus)
In calculus, a differential is traditionally an infinitesimally small change in a variable. For example, if x is a variable, then a change in the value of x is often denoted Δx . The differential dx represents such a change, but is infinitely small...

. Here is Leibniz's argument: Let u(x) and v(x) be two differentiable function
Differentiable function
In calculus , a differentiable function is a function whose derivative exists at each point in its domain. The graph of a differentiable function must have a non-vertical tangent line at each point in its domain...

s of x. Then the differential of uv is NEWLINE
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NEWLINE \begin{align} y + d(u\cdot v) & {} = (u + du)\cdot (v + dv) \\ d(u\cdot v) & {} = (u + du)\cdot (v + dv) - u\cdot v \\ & {} = u\cdot dv + v\cdot du + du\cdot dv. \end{align} Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that $d\left(u\cdot v\right) = v\cdot du + u\cdot dv \,\!$ and this is indeed the differential form of the product rule. If we divide through by the differential dx, we obtain $\frac\left\{d\right\}\left\{dx\right\} \left(u\cdot v\right) = v \cdot \frac\left\{du\right\}\left\{dx\right\} + u \cdot \frac\left\{dv\right\}\left\{dx\right\} \,\!$ which can also be written in "prime notation" as $\left(u\cdot v\right)\text{'} = v\cdot u\text{'} + u\cdot v\text{'}. \,\!$

## Examples

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• Suppose we want to differentiate ƒ(x) = x2 sin
Sine
In mathematics, the sine function is a function of an angle. In a right triangle, sine gives the ratio of the length of the side opposite to an angle to the length of the hypotenuse.Sine is usually listed first amongst the trigonometric functions....

(x). By using the product rule, one gets the derivative ƒ '(x) = 2x sin(x) + x2cos(x) (since the derivative of x2 is 2x and the derivative of sin(x) is cos(x)).
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• One special case of the product rule is the constant multiple rule which states: if c is a real number
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...

and ƒ(x) is a differentiable function, then (x) is also differentiable, and its derivative is (c × ƒ)'(x) = c × ƒ '(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
Linear transformation
In mathematics, a linear map, linear mapping, linear transformation, or linear operator is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. As a result, it always maps straight lines to straight lines or 0...

.
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• The rule for integration by parts
Integration by parts
In calculus, and more generally in mathematical analysis, integration by parts is a rule that transforms the integral of products of functions into other integrals...

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 if it is differentiable.)
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## A common error

It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u ′)(v ′). Leibniz himself made this error initially; however, there are clear counterexample
Counterexample
In logic, and especially in its applications to mathematics and philosophy, a counterexample is an exception to a proposed general rule. For example, consider the proposition "all students are lazy"....

s. Consider a differentiable function ƒ(x) whose derivative is ƒ '(x). This function can also be written as ƒ(x) · 1, since 1 is the identity element
Identity element
In mathematics, an identity element is a special type of element of a set with respect to a binary operation on that set. It leaves other elements unchanged when combined with them...

for multiplication. If the above-mentioned misconception were true, (u′)(v′) would equal zero
0 (number)
0 is both a numberand the numerical digit used to represent that number in numerals.It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems...

. This is true because the derivative of a constant
Derivative of a constant
In calculus, the derivative of a constant function is zero .The rule can be justified in various ways...

(such as 1) is zero and the product of ƒ '(x) · 0 is also zero.

## Proof of the product rule

A rigorous proof of the product rule can be given using the properties of limits
Limit (mathematics)
In mathematics, the concept of a "limit" is used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concept of limit allows mathematicians to define a new point from a Cauchy sequence of previously defined points within a complete metric...

and the definition of the derivative as a limit of Newton
Isaac Newton
Sir Isaac Newton PRS was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian, who has been "considered by many to be the greatest and most influential scientist who ever lived."...

's difference quotient
Difference quotient
The primary vehicle of calculus and other higher mathematics is the function. Its "input value" is its argument, usually a point expressible on a graph...

. If $h\left(x\right) = f\left(x\right)g\left(x\right),\,$ and ƒ and g are each differentiable at the fixed number x, then $h\text{'}\left(x\right) = \lim_\left\{w\to x\right\}\left\{ h\left(w\right) - h\left(x\right) \over w - x\right\} = \lim_\left\{w\to x\right\}\left\{f\left(w\right)g\left(w\right) - f\left(x\right)g\left(x\right) \over w - x\right\}. \qquad\qquad\left(1\right)$ Now the difference $f\left(w\right)g\left(w\right) - f\left(x\right)g\left(x\right)\qquad\qquad\left(2\right)$ is the area of the big rectangle minus the area of the small rectangle in the illustration. The region between the smaller and larger rectangle can be split into two rectangles, the sum of whose areas is $f\left(x\right) \Bigg\left( g\left(w\right) - g\left(x\right) \Bigg\right) + g\left(w\right)\Bigg\left( f\left(w\right) - f\left(x\right) \Bigg\right).\qquad\qquad\left(3\right)$ Therefore the expression in (1) is equal to $\lim_\left\{w\to x\right\}\left\left( f\left(x\right) \left\left( \left\{g\left(w\right) - g\left(x\right) \over w - x\right\} \right\right) + g\left(w\right)\left\left( \left\{f\left(w\right) - f\left(x\right) \over w - x\right\} \right\right) \right\right).\qquad\qquad\left(4\right)$ Assuming that all limits used exist, (4) is equal to $\left\left(\lim_\left\{w\to x\right\}f\left(x\right)\right\right) \left\left(\lim_\left\{w\to x\right\} \left\{g\left(w\right) - g\left(x\right) \over w - x\right\}\right\right) + \left\left(\lim_\left\{w\to x\right\} g\left(w\right)\right\right) \left\left(\lim_\left\{w\to x\right\} \left\{f\left(w\right) - f\left(x\right) \over w - x\right\} \right\right). \qquad\qquad\left(5\right)$ Now $\lim_\left\{w\to x\right\}f\left(x\right) = f\left(x\right)$ This holds because f(x) remains constant as w → x. $\lim_\left\{w\to x\right\} g\left(w\right) = g\left(x\right)\,$ This holds because differentiable functions are continuous (g is assumed differentiable in the statement of the product rule). Also: $\lim_\left\{w\to x\right\} \left\{f\left(w\right) - f\left(x\right) \over w - x\right\} = f\text{'}\left(x\right)$    and    $\lim_\left\{w\to x\right\} \left\{g\left(w\right) - g\left(x\right) \over w - x\right\} = g\text{'}\left(x\right)$ because f and g are differentiable at x; We conclude that the expression in (5) is equal to $f\left(x\right)g\text{'}\left(x\right) + g\left(x\right)f\text{'}\left(x\right). \,$

### A Brief Proof

By definition, if $f, g: \mathbb\left\{R\right\} \rightarrow \mathbb\left\{R\right\}$ are differentiable at $x_0$ then we can write$f\left(x+h\right) = f\left(x\right) + f\text{'}\left(x\right)h + \psi_1\left(h\right) \qquad \qquad g\left(x+h\right) = g\left(x\right) + g\text{'}\left(x\right)h + \psi_2\left(h\right)$ such that $\lim_\left\{h \to 0\right\} \frac\left\{\psi_1\left(h\right)\right\}\left\{h\right\} = \lim_\left\{h \to 0\right\} \frac\left\{\psi_2\left(h\right)\right\}\left\{h\right\} = 0$, that is, $\psi_1, \psi_2 \sim O\left(h\right)$. Then: \begin\left\{align\right\} fg\left(x+h\right) - fg\left(x\right) = \left(f\left(x\right) + f\text{'}\left(x\right)h +\psi_1\left(h\right)\right)\left(g\left(x\right) + g\text{'}\left(x\right)h + \psi_2\left(h\right)\right) - fg\left(x\right)= f\text{'}\left(x\right)g\left(x\right)h + f\left(x\right)g\text{'}\left(x\right)h + O\left(h\right) \\\left[12pt\right] \end\left\{align\right\} Taking the limit for small $h$ gives the result.

### Using logarithms

Let f = uv and suppose u and v are positive functions of x. Then $\ln f =\ln \left(u\cdot v\right)=\ln u + \ln v.\,$ Differentiating both sides: $\left\{1 \over f\right\} \left\{df \over dx\right\} = \left\{1 \over u\right\} \left\{du \over dx\right\} + \left\{1 \over v\right\} \left\{dv \over dx\right\}\,$ and so, multiplying the left side by f, and the right side by uv, $\left\{df \over dx\right\} = v \left\{du \over dx\right\} + u \left\{dv \over dx\right\}.\,$ The proof appears in http://planetmath.org/encyclopedia/LogarithmicProofOfProductRule.html. Note that since u, v need to be continuous, the assumption on positivity does not diminish the generality. This proof relies on the chain rule
Chain rule
In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function in terms of the derivatives of f and g.In integration, the...

and on the properties of the natural logarithm
Natural logarithm
The natural logarithm is the logarithm to the base e, where e is an irrational and transcendental constant approximately equal to 2.718281828...

function, both of which are deeper than the product rule. From one point of view, that is a disadvantage of this proof. On the other hand, the simplicity of the algebra in this proof perhaps makes it easier to understand than a proof using the definition of differentiation directly.

### Using the chain rule

The product rule can be considered a special case of the chain rule
Chain rule
In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function in terms of the derivatives of f and g.In integration, the...

for several variables. NEWLINE
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$\left\{d \left(ab\right) \over dx\right\} = \frac\left\{\partial\left(ab\right)\right\}\left\{\partial a\right\}\frac\left\{da\right\}\left\{dx\right\}+\frac\left\{\partial \left(ab\right)\right\}\left\{\partial b\right\}\frac\left\{db\right\}\left\{dx\right\} = b \frac\left\{da\right\}\left\{dx\right\} + a \frac\left\{db\right\}\left\{dx\right\}. \,$
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### Using non-standard analysis

Let u and v be continuous functions in x, and let dx, du and dv be infinitesimal
Infinitesimal
Infinitesimals have been used to express the idea of objects so small that there is no way to see them or to measure them. The word infinitesimal comes from a 17th century Modern Latin coinage infinitesimus, which originally referred to the "infinite-th" item in a series.In common speech, an...

s within the framework of non-standard analysis
Non-standard analysis
Non-standard analysis is a branch of mathematics that formulates analysis using a rigorous notion of an infinitesimal number.Non-standard analysis was introduced in the early 1960s by the mathematician Abraham Robinson. He wrote:...

, specifically the hyperreal number
Hyperreal number
The system of hyperreal numbers represents a rigorous method of treating the infinite and infinitesimal quantities. The hyperreals, or nonstandard reals, *R, are an extension of the real numbers R that contains numbers greater than anything of the form1 + 1 + \cdots + 1. \, Such a number is...

s. Using st to denote the standard part function
Standard part function
In non-standard analysis, the standard part function is a function from the limited hyperreals to the reals, which associates to every hyperreal, the unique real infinitely close to it. As such, it is a mathematical implementation of the historical concept of adequality introduced by Pierre de...

that associates to a finite hyperreal number the real infinitely close to it, this gives NEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINENEWLINE
 $\frac\left\{d\left(uv\right)\right\}\left\{dx\right\}\,$ $=\operatorname\left\{st\right\}\left\left(\frac\left\{\left(u + \mathrm du\right)\left(v + \mathrm dv\right) - uv\right\}\left\{\mathrm dx\right\}\right\right)$ $=\operatorname\left\{st\right\}\left\left(\frac\left\{uv + u \cdot \mathrm dv + v \cdot \mathrm du + \mathrm dv \cdot \mathrm du -uv\right\}\left\{\mathrm dx\right\}\right\right)$ $=\operatorname\left\{st\right\}\left\left(\frac\left\{u \cdot \mathrm dv + \left(v + \mathrm dv\right) \cdot \mathrm du\right\}\left\{\mathrm dx\right\}\right\right)$ $=\left\{u\right\}\frac\left\{dv\right\}\left\{dx\right\} + \left\{v\right\}\frac\left\{du\right\}\left\{dx\right\}$
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### Using smooth infinitesimal analysis

In the context of Lawvere's approach to infinitesimals, let du and dv be nilsquare infinitesimals. Then NEWLINE
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NEWLINE \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} provided that $du \cdot dv = 0\,\!$ (this may not actually be true even for nilsquare infinitesimals in general).

### 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 $\frac\left\{d\left(uvw\right)\right\}\left\{dx\right\} = \frac\left\{du\right\}\left\{dx\right\}vw + u\frac\left\{dv\right\}\left\{dx\right\}w + uv\frac\left\{dw\right\}\left\{dx\right\}\,\!$. For a collection of functions $f_1, \dots, f_k$, we have $\frac\left\{d\right\}\left\{dx\right\} \left \left[ \prod_\left\{i=1\right\}^k f_i\left(x\right) \right \right] = \sum_\left\{i=1\right\}^k \left\left(\frac\left\{d\right\}\left\{dx\right\} f_i\left(x\right) \prod_\left\{j\ne i\right\} f_j\left(x\right) \right\right)$

### Higher derivatives

It can also be generalized to the Leibniz rule
Leibniz rule (generalized product rule)
In calculus, the general Leibniz rule, named after Gottfried Leibniz, generalizes the product rule. It states that if f and g are n-times differentiable functions, then the nth derivative of the product fg is given by...

for the nth derivative of a product of two factors:$\left(uv\right)^\left\{\left(n\right)\right\}\left(x\right) = \sum_\left\{k=0\right\}^n \left\{n \choose k\right\} \cdot u^\left\{\left(n-k\right)\right\}\left(x\right)\cdot v^\left\{\left(k\right)\right\}\left(x\right).$ See also binomial coefficient
Binomial coefficient
In mathematics, binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. They are indexed by two nonnegative integers; the binomial coefficient indexed by n and k is usually written \tbinom nk , and it is the coefficient of the x k term in...

and the formally quite similar binomial theorem
Binomial theorem
In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with , and the coefficient a of...

Leibniz rule (generalized product rule)
In calculus, the general Leibniz rule, named after Gottfried Leibniz, generalizes the product rule. It states that if f and g are n-times differentiable functions, then the nth derivative of the product fg is given by...

.

### Higher partial derivatives

For partial derivative
Partial derivative
In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant...

s, we have $\left\{\partial^n \over \partial x_1\,\cdots\,\partial x_n\right\} \left(uv\right) \sum_S \left\{\partial^\left\{|S|\right\} u \over \prod_\left\{i\in S\right\} \partial x_i\right\} \cdot \left\{\partial^\left\{n-|S|\right\} v \over \prod_\left\{i\not\in S\right\} \partial x_i\right\}$ where the index S runs through the whole list of 2n subsets of {1, ..., n}. For example, when n = 3, then \begin\left\{align\right\} &\left\{\right\}\quad \left\{\partial^3 \over \partial x_1\,\partial x_2\,\partial x_3\right\} \left(uv\right) \\ \\ &\left\{\right\}= u \cdot\left\{\partial^3 v \over \partial x_1\,\partial x_2\,\partial x_3\right\} + \left\{\partial u \over \partial x_1\right\}\cdot\left\{\partial^2 v \over \partial x_2\,\partial x_3\right\} + \left\{\partial u \over \partial x_2\right\}\cdot\left\{\partial^2 v \over \partial x_1\,\partial x_3\right\} + \left\{\partial u \over \partial x_3\right\}\cdot\left\{\partial^2 v \over \partial x_1\,\partial x_2\right\} \\ \\ &\left\{\right\}\qquad + \left\{\partial^2 u \over \partial x_1\,\partial x_2\right\}\cdot\left\{\partial v \over \partial x_3\right\} + \left\{\partial^2 u \over \partial x_1\,\partial x_3\right\}\cdot\left\{\partial v \over \partial x_2\right\} + \left\{\partial^2 u \over \partial x_2\,\partial x_3\right\}\cdot\left\{\partial v \over \partial x_1\right\} + \left\{\partial^3 u \over \partial x_1\,\partial x_2\,\partial x_3\right\}\cdot v. \end\left\{align\right\}

### A product rule in Banach spaces

Suppose X, Y, and Z are Banach space
Banach space
In mathematics, Banach spaces is the name for complete normed vector spaces, one of the central objects of study in functional analysis. A complete normed vector space is a vector space V with a norm ||·|| such that every Cauchy sequence in V has a limit in V In mathematics, Banach spaces is the...

s (which includes Euclidean space
Euclidean space
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...

) and B : X × YZ is a continuous bilinear operator
Bilinear operator
In mathematics, a bilinear operator is a function combining elements of two vector spaces to yield an element of a third vector space that is linear in each of its arguments. Matrix multiplication is an example.-Definition:...

. Then B is differentiable, and its derivative at the point (x,y) in X × Y is the linear map D(x,y)B : X × YZ given by$\left(D_\left\left( x,y \right\right)\,B\right)\left\left( u,v \right\right) = B\left\left( u,y \right\right) + B\left\left( x,v \right\right)\qquad\forall \left(u,v\right)\in X \times Y.$

### Derivations in abstract algebra

In abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...

, the product rule is used to define what is called a derivation
Derivation (abstract algebra)
In abstract algebra, a derivation is a function on an algebra which generalizes certain features of the derivative operator. Specifically, given an algebra A over a ring or a field K, a K-derivation is a K-linear map D: A → A that satisfies Leibniz's law: D = b + a.More...

, not vice versa.

### For vector functions

The product rule extends to scalar multiplication
Scalar multiplication
In mathematics, scalar multiplication is one of the basic operations defining a vector space in linear algebra . In an intuitive geometrical context, scalar multiplication of a real Euclidean vector by a positive real number multiplies the magnitude of the vector without changing its direction...

, dot product
Dot product
In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number obtained by multiplying corresponding entries and then summing those products...

s, and cross product
Cross product
In mathematics, the cross product, vector product, or Gibbs vector product is a binary operation on two vectors in three-dimensional space. It results in a vector which is perpendicular to both of the vectors being multiplied and normal to the plane containing them...

s of vector functions. For scalar multiplication: $\left(f \cdot \vec g\right)\text{'} = f\;\text{'}\cdot \vec g + f \cdot \vec g\;\text{'} \,$ For dot products: $\left(\vec f \cdot \vec g\right)\text{'} = \vec f\;\text{'}\cdot \vec g + \vec f \cdot \vec g\;\text{'} \,$ For cross products: $\left(\vec f \times \vec g\right)\text{'} = \vec f\;\text{'} \times \vec g + \vec f \times \vec g\;\text{'} \,$ (Beware: since cross products are not commutative, it is not correct to write $\left(f\times g\right)\text{'}=f\text{'}\times g+g\text{'}\times f. \,$ But cross products are anticommutative, so it can be written as $\left(f\times g\right)\text{'}=f\text{'}\times g-g\text{'}\times f. \,$

### For scalar fields

For scalar fields the concept of gradient
In vector calculus, the gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change....

is the analog of the derivative: $\nabla \left(f \cdot g\right) = \nabla f \cdot g + f \cdot \nabla g \,$

## An application

Among the applications of the product rule is a proof that $\left\{d \over dx\right\} x^n = nx^\left\{n-1\right\}\,\!$ 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
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers...

on the exponent n. If n = 0 then xn is constant and nxn − 1 = 0. The rule holds in that case because the derivative of a constant function is 0. If the rule holds for any particular exponent n, then for the next value, n + 1, we have \begin\left\{align\right\} \left\{d \over dx\right\}x^\left\{n+1\right\} &\left\{\right\}= \left\{d \over dx\right\}\left\left( x^n\cdot x\right\right) \\\left[12pt\right] &\left\{\right\}= x\left\{d \over dx\right\} x^n + x^n\left\{d \over dx\right\}x \qquad\mbox\left\{\left(the product rule is used here\right)\right\} \\\left[12pt\right] &\left\{\right\}= x\left\left(nx^\left\{n-1\right\}\right\right) + x^n\cdot 1\qquad\mbox\left\{\left(the induction hypothesis is used here\right)\right\} \\\left[12pt\right] &\left\{\right\}= \left(n + 1\right)x^n. \end\left\{align\right\} Therefore if the proposition is true of n, it is true also of n + 1.

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• General Leibniz rule
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• Reciprocal rule
Reciprocal rule
In calculus, the reciprocal rule is a shorthand method of finding the derivative of a function that is the reciprocal of a differentiable function, without using the quotient rule or chain rule....

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• Differential (calculus)
Differential (calculus)
In calculus, a differential is traditionally an infinitesimally small change in a variable. For example, if x is a variable, then a change in the value of x is often denoted Δx . The differential dx represents such a change, but is infinitely small...

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• Derivation (abstract algebra)
Derivation (abstract algebra)
In abstract algebra, a derivation is a function on an algebra which generalizes certain features of the derivative operator. Specifically, given an algebra A over a ring or a field K, a K-derivation is a K-linear map D: A → A that satisfies Leibniz's law: D = b + a.More...

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• Product Rule Practice Problems [Kouba, University of California: Davis]
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