Product Rule (calculus)

Discovery of this rule is credited to Leibniz, 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" (i.e. at least quadratic in 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

For example, suppose you want to differentiate f(x) = x² sin(x). By using the product rule, you get the derivative 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 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.

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 also be written as f(x) · 1, since 1 is the identity element for multiplication. 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 the zero function; such a belief is not correct.

Proof of the product rule

Suppose

and suppose further that g and h are each differentiable at the fixed number x. Then

Since

,

we have

Since h is continuous at x, we have

and by the definition of the derivative, and the differentiability of h and g at x, we also have

Thus, we are justified in splitting each of the products inside the limit, and putting everything together, we have

and this completes the proof.

Generalizations

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

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

(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 × Y → Z 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 × Y → Z given by

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

See also:

• Quotient Rule

Referenced By