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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
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.
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2 Proof of the product rule 3 Generalizations |
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.
Suppose
The product rule can be generalised to products of more than two factors. For example, for three factors we have
In multivariable calculus, the product rule is also valid for different notions of "product": scalar product and cross product of vectorss, 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
Common misconception
Proof of the product rule
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 quotientss:
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
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.
See also: