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or in Leibniz's notation
In differentials notation, this can be written as
In Leibniz's notation, the derivative of the product of three functions (not to be confused with Euler's triple product rule) is
Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials. (However, J. M. Child, a translator of Leibniz's papers, 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
- Suppose we want to differentiate f(x) = x2 sin(x). By using the product rule, one gets the derivative f′(x) = 2x sin(x) + x2 cos(x) (since the derivative of x2 is 2x and the derivative of the sine function is the cosine function).
- One special case of the product rule is the constant multiple rule, which states: if c is a number and f(x) is a differentiable function, then cf(x) is also differentiable, and its derivative is (cf)′(x) = cf′(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.)
Proof by factoring (from first principles)
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). To do this, (which is zero, and thus does not change the value) is added to the numerator to permit its factoring, and then properties of limits are used.
The fact that
is deduced from a theorem that states that differentiable functions are continuous.
By definition, if are differentiable at then we can write
such that also written . Then:
Taking the limit for small gives the result.
Differentiating both sides:
The product rule can be considered a special case of the chain rule for several variables.
Let u and v be continuous functions in x, and let dx, du and dv 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
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
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
Applied at a specific point x, the above formula gives:
Furthermore, for the nth derivative of an arbitrary number of factors:
Higher partial derivatives
For partial derivatives, we have
Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B : X × Y → Z 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 × Y → Z given by
Derivations in abstract algebra
For scalar multiplication:
For dot products:
For cross products:
Note: cross products are not commutative, i.e. , instead products are anticommutative, so it can be written as
For scalar fields the concept of gradient is the analog of the derivative:
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 for n, it is true also for n + 1, and therefore for all natural n.