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In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated thus:

or in the Leibniz notation thus:

.

The derivative of the product of three functions is:

.

Since the product of two or more functions occurs in many mathematical models of physical phenomena, the product rule has broad application in Physics, Chemistry, and Engineering.

Discovery by Leibniz

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Discovery of this rule is credited to Gottfried Leibniz (however, Child (2008) argues that it is due to Isaac Barrow), who demonstrated it using differentials. Here is Leibniz's argument: Let u(x) and v(x) be two differentiable functions of x. Then the differential of uv is

 

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

 

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

 

which can also be written in "prime notation" as

 

Examples

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  • Suppose one wants to differentiate ƒ(x) = x2 sin(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)).
  • One special case of the product rule is the constant multiple rule which states: if c is a real number 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.
  • 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 if it is differentiable.)

Physics Example I: Rocket Acceleration

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The acceleration of model rockets can be computed with the product rule.

Consider the vertical acceleration of a model rocket relative to its initial position at a fixed point on the ground. Newton's second law says that the force is equal to the time rate change of momentum. If F is the net force (sum of forces), p is the momentum, and t is the time,

 

Since the momentum is equal to the product of mass and velocity, this yields

 

where m is the mass and v is the velocity. Application of the product rule gives

 

Since the acceleration a, is defined as the time rate change of velocity, a = dv/dt,

 

Solving for the acceleration,

 

Since the rocket is losing mass, dm/dt is negative, and the changing mass term results in increased acceleration.[1][2]

Physics Example II: Electromagnetic induction

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Faraday's law of electromagnetic induction states that the induced electromotive force is the negative time rate of change of magnetic flux through a conducting loop.

 

where   is the electromotive force (emf) in volts and ΦB is the magnetic flux in webers. For a loop of area, A, in a magnetic field, B, the magnetic flux is given by

 

where θ is the angle between the normal to the current loop and the magnetic field direction.

Taking the negative derivative of the flux with respect to time yields the electromotive force gives

 

In many cases of practical interest, only one variable (A, B, or θ) is changing so two of the three above terms are often zero.


A common error

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It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u ′)(v ′) (there is an exaggerated story that Leibniz himself made this error initially);[3] however, there are clear counterexamples to this. For a ƒ(x) whose derivative is ƒ '(x), the function can also be written as ƒ(x) · 1, since 1 is the identity element for multiplication. If the above-mentioned misconception were true, (u′)(v′) would equal zero. This is true because the derivative of a constant (such as 1) is zero and the product of ƒ '(x) · 0 is also zero.

Proof of the product rule

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A rigorous proof of the product rule can be given using the properties of limits and the definition of the derivative as a limit of Newton's difference quotient.

If

 

and ƒ and g are each differentiable at the fixed number x, then

 

Now the difference

 

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[4]

 

Therefore the expression in (1) is equal to

 

Assuming that all limits used exist, (4) is equal to

 

Now

 

because ƒ(x) remains constant as wx;

 

because g is differentiable at x;

 

because ƒ is differentiable at x;

and now the "hard" one:

 

because g, being differentiable, is continuous at x.

We conclude that the expression in (5) is equal to

 

Alternative proof

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Suppose : 

By applying Newton's difference quotient and the limit as h approaches 0, we are able to represent the derivative in the form

 

In order to simplify this limit we add and subtract the term   to the numerator, keeping the fraction's value unchanged

 

This allows us to factorise the numerator like so

 

The fraction is split into two

 

The limit is applied to each term and factor of the limit expression

 

Each limit is evaluated. Taking into consideration the definition of the derivative, the result is

 

Using logarithms

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Let f = uv and suppose u and v are positive functions of x. Then

 

Differentiating both sides:

 

and so, multiplying the left side by f, and the right side by uv,

 

The proof appears in [1]. 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 and on the properties of the natural logarithm 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

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

 

Using non-standard analysis

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Let u and v be continuous functions in x, and let dx, du and dv be infinitesimals. This gives,


   
 
 
 

Generalizations

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A product of more than two factors

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

 .

For a collection of functions  , we have

 

Higher derivatives

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It can also be generalized to the Leibniz rule for the nth derivative of a product of two factors:

 

See also binomial coefficient and the formally quite similar binomial theorem. See also Leibniz rule (generalized product rule).

Higher partial derivatives

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For partial derivatives, we have

 

where the index S runs through the whole list of 2n subsets of {1, ..., n}. If this seems hard to understand, consider the case in which n = 3:

 

A product rule in Banach spaces

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Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B : X × YZ is a continuous bilinear operator. 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

 

Derivations in abstract algebra

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In abstract algebra, the product rule is used to define what is called a derivation, not vice versa.

For vector functions

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The product rule extends to scalar multiplication, dot products, and cross products of vector functions.

For scalar multiplication:  

For dot products:  

For cross products:  

(Beware: since cross products are not commutative, it is not correct to write   But cross products are anticommutative, so it can be written as  )

For scalar fields

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For scalar fields the concept of gradient is the analog of the derivative:

 

An application

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Among the applications of the product rule is a proof that

 

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 = 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

 

Therefore if the proposition is true of n, it is true also of n + 1.

See also

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References

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  1. ^ Newton’s Second Law for Systems with Variable Mass, David Chandler, The Physics Teacher -- October 2000 -- Volume 38, Issue 7, pp. 396
  2. ^ Measuring thrust and predicting trajectory in model rocketry,Michael Courtney and Amy Courtney, arXiv:0903.1555
  3. ^ Michelle Cirillo (August 2007). "Humanizing Calculus" (PDF). The Mathematics Teacher. 101 (1): 23–27. doi:10.5951/MT.101.1.0023.{{cite journal}}: CS1 maint: date and year (link)
  4. ^ The illustration disagrees with some special cases, since – in actuality – ƒ(w) need not be greater than ƒ(x) and g(w) need not be greater than g(x). Nonetheless, the equality of (2) and (3) is easily checked by algebra.
  • Child, J. M. (2008) "The early mathematical manuscripts of Leibniz", Gottfried Wilhelm Leibniz, translated by J. M. Child; page 29, footnote 58.

Category:Differentiation rules Category:Articles containing proofs