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Product rule

The product rule, of calculus, governs the differentiation of products of differentiable functions. Development of this rule is credited to Leibniz; who demonstrated that (x + dx)(y + dy) - xy = x(dy) + y(dx); as (dx)(dy) is "negligible". If two differentiable functions u and v of the variable x are given, then the derivative of their product uv equals the derivative of u, multipled by v; and added to u, multiplied by the derivative of v:

<math>\frac{d}{dx} (uv) = \left( \frac{du}{dx} \right) v + u \left( \frac{dv}{dx} \right)</math>

or as a statement about differentials:

<math>d(uv) = (du)v + u(dv)\,</math>

or in alternative notation

<math>(uv)' = u' v + u v'\,</math>

For example, suppose you want to differentiate f(x) = x² sin(x). By using the product rule, you get the derivative 2x sin(x) + x² cos(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 f(x) is a differentiable function, then cf(x) is also differentiable and it derivative equals c multiplied by the derivative of f(x). (This follows from the product rule since the derivative of any constant is zero.)

The product rule can be used to derive the rule for integration by parts and the quotient rule.

Table of contents

Common misconception

It is a common misconception, when studying calculus, to suppose that the derivative of (uv) equals (u')(v'); however, it is quite easy to find counterexamples to this. Most simply, take a function f, whose derivative is f '(x). Now that function can be also written f(x) · 1, since 1 is the identity element for multiplication. Now, suppose the above mentioned misconception were true; if so, (u')(v') would equal zero; since, the derivative of a constant (such as 1) is zero; and, the product, of any number and zero, is zero. Such a misconception could result in a belief that the derivative, of every function, is 0; such a belief is not correct.

Informal Proof A proof of this rule can be derived from Newton's difference quotient: The derivative of [f(x)][g(x)] = (the limit as h approaches 0) {[f(x + h)g(x + h)] - [f(x)g(x)]} / h; for which a sum is added and subtracted (equivalent to adding 0, an algebraic manipulation): {[f(x + h)g(x + h)] - [f(x)g(x)]} + {−[f(x + h)g(x)] + f(x + h)g(x)]} / h; which is then reduced and broken into two sections: f(f + h){[g(x) + h) - g(x)] / h} + g(x){[f(x + h) - f(x)] / h}; which equals (as h approaches 0): the limit of f(x + h) · the limit of {[g(x) + h) - g(x)] / h} + the limit of g(x) · the limit of {[f(x + h) - f(x)] / h}; which equals: f(x)g'(x) + g(x)f'(x).

Informal justification of the product rule

The product rule can be justified as follows. Let:

<math>y = uv</math>

and suppose that you want to find dy/dx.

Let the independent variable x increase by a small amount Δx, resulting in an increase of u by Δu, of v by Δv, and of y by Δy. Now:

<math>y + \Delta y = (u + \Delta u)(v + \Delta v)</math>

Multiply out to obtain:

<math>y + \Delta y = uv + u \Delta v + v \Delta u + \Delta u \Delta v = y + u \Delta v + v \Delta u + \Delta u \Delta v</math>

So, cancelling y from the left and right hand sides,

<math>\Delta y = u \Delta v + v \Delta u + \Delta u \Delta v</math>

Now divide throughout by Δx:

<math>\frac{\Delta y}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}</math>

Let Δx tend to 0, then Δu tends to 0 also, since u is a differentiable and hence continuous function of x. We get:

<math>\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} + 0\frac{dv}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}</math>

which proves the product rule.

Generalizations

The product rule can be generalised 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>

It can also be generalized to higher derivatives of products of two factors: if y = uv and y(n) denotes the n-th derivative of y, then

<math>y^{(n)}(x) = \sum_{k=0}^n {n \choose k} u^{(n-k)}(x)\; v^{(k)}(x)</math>

(see binomial coefficient). This result is formally quite similar to the binomial theorem.

In multivariable calculus, the product rule is also valid for different notions of "product": scalar product and cross product of vectors, matrix product, inner products etc. All of these are summarized by the following general statement: let X, Y, Z be Banach spaces (which includes Euclidean space) and let B : X × YZ be 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

D(x,y)B(u,v) = B(x,v) + B(u,y)   for all (u,v) in X × Y.

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