# Product rule

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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 as

or in the Leibniz notation

In the notation of differentials this can be written as

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

## Discovery

Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials.^{[1]} (However, Child (2008) argues that it is due to Isaac Barrow). 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 Lagrange's notation as

## Examples

- Suppose we want to differentiate
*ƒ*(*x*) =*x*^{2}sin(*x*). By using the product rule, one gets the derivative*ƒ*'(*x*) = 2*x*sin(*x*) +*x*^{2}cos(*x*) (since the derivative of*x*^{2}is 2*x*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*cƒ*(*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.)

## Proofs

### Simple Proof

Let *h(x) = f(x) g(x)*, and suppose that *f* and *g* are each differentiable at *x*. We want to prove that *h* is differentiable at *x* and that its derivative *h'(x)* is given by *f'(x) g(x) + f(x) g'(x)*.

### More Complicated Proof

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Let *h(x) = f(x) g(x)*, and suppose that *f* and *g* are each differentiable at *x _{0}*. (Note that

*x*will remain fixed throughout the proof). We want to prove that

_{0}*h*is differentiable at

*x*and that its derivative

_{0}*h'(x*is given by

_{0})*f'(x*.

_{0}) g(x_{0}) + f(x_{0}) g'(x_{0})Let *Δh = h(x _{0}+Δx) - h(x_{0})*; note that although

*x*is fixed,

_{0}*Δh*depends on the value of

*Δx*, which is thought of as being "small."

The function *h* is differentiable at *x _{0}* if the limit

exists; when it does, *h'(x _{0})* is defined to be the value of the limit.

As with *Δh*, let *Δf = f(x _{0}+Δx) - f(x_{0})* and

*Δg = g(x*which, like

_{0}+Δx) - g(x_{0})*Δh*, also depends on

*Δx*. Then

*f(x*and

_{0}+Δx) = f(x_{0}) + Δf*g(x*.

_{0}+Δx) = g(x_{0}) + ΔgIt follows that *h(x _{0}+Δx) = f(x_{0}+Δx) g(x_{0}+Δx) = (f(x_{0}) + Δf) (g(x_{0})+Δg)*; applying the distributive law, we see that

While it is not necessary for the proof, it can be helpful to understand this product geometrically as the area of the rectangle in this diagram:

To get the value of *Δh*, subtract *h(x _{0})=f(x_{0}) g(x_{0})* from equation Template:EquationNote. This removes the area of the white rectangle, leaving three rectangles:

To find *h'(x _{0})*, we need to find the limit as

*Δx*goes to 0 of

The first two terms of the right-hand side of this equation correspond to the areas of the blue rectangles; the third corresponds to the area of the gray rectangle. Using the basic properties of limits and the definition of the derivative, we can tackle this term-by term. First,

Similarly,

The third term, corresponding to the small gray rectangle, winds up being negligible (i.e. going to 0 in the limit) because *Δf Δg* "vanishes to second order." Rigorously,

We have shown that the limit of each of the three terms on the right-hand side of equation Template:EquationNote exists, hence

exists and is equal to the sum of the three limits. Thus, the product *h(x)* is differentiable at *x _{0}* and its derivative is given by

as was to be shown.

### Brief proof

By definition, if are differentiable at then we can write

such that , also written . Then:

Taking the limit for small gives the result.

### Logarithms and quarter squares

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* (note: *f* = *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 (however, information about the derivative of a logarithm that is sufficient to carry out a variant of the proof can be inferred by considering the derivative at *x* = *1* of the logarithm to any base of *cx*, where *c* is a constant, then generalising *c*). 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.

There is an analogous but arguably even easier proof (i.e., some people may find it easier as it can be used before being able to differentiate logarithms), using quarter square multiplication, which similarly relies on the chain rule and on the properties of the quarter square function (shown here as *q*, i.e., with ):

Differentiating both sides:

This does not present issues of whether the values are positive or negative, and the function's properties are much simpler to demonstrate (indeed, it can be differentiated without using first principles by considering the derivative at *x* = *0* of *cx*, where *c* is a constant, then generalising *c*).

Note also, these proofs are only valid for numbers or similar, whereas proofs from first principles are also valid for matrices and such like.

### Chain rule

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

### Non-standard analysis

Let *u* and *v* be continuous functions in *x*, and let d*x*, d*u* and d*v* be infinitesimals within the framework of non-standard analysis, specifically the hyperreal numbers. Using st to denote the standard part function that associates to a finite hyperreal number the real infinitely close to it, this gives

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 = u' dx and dv = v' dx, so that

since

## Generalizations

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### 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

For a collection of functions , we have

### Higher derivatives

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

See also binomial coefficient and the formally quite similar binomial theorem. See also General Leibniz rule.

Furthermore, for the *n*th derivative of an arbitrary number of factors:

### Higher partial derivatives

For partial derivatives, we have

where the index *S* runs through the whole list of 2^{n} subsets of {1, ..., *n*}. For example, when *n* = 3, then