Suppose we want to differentiate ƒ(x) = x2sin(x). By using the product rule, one gets the derivative ƒ '(x) = 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 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.)
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.
Let h(x) = f(x) g(x), and suppose that f and g are each differentiable at x0. (Note that x0 will remain fixed throughout the proof). We want to prove that h is differentiable at x0 and that its derivative h′(x0) is given by f′(x0) g(x0) + f(x0) g′(x0).
Let Δh = h(x0 + Δx) − h(x0); note that although x0 is fixed, Δh depends on the value of Δx, which is thought of as being "small".
The function h is differentiable at x0 if the limit
exists; when it does, h′(x0) is defined to be the value of the limit.
As with Δh, let Δf = f(x0 + Δx) − f(x0) and Δg = g(x0 + Δx) − g(x0) which, like Δh, also depends on Δx. Then f(x0 + Δx) = f(x0) + Δf and g(x0 + Δx) = g(x0) + Δg.
It follows that h(x0 + Δx) = f(x0 + Δx) g(x0 + Δx) = (f(x0) + Δf) (g(x0) + 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(x0) = f(x0) g(x0) from equation (*). This removes the area of the white rectangle, leaving three rectangles:
To find h′(x0), 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,
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 (**) exists, hence
exists and is equal to the sum of the three limits. Thus, the product h(x) is differentiable at x0 and its derivative is given by
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 . 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.
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 of n, it is true also of n + 1, and therefore all integral n.
Product rule is also used in definition of abstract tangent space of some abstract geometric figure (smooth manifold). This definition we can use if we cannot or wish to not use surrounding ambient space where our chosen geometric figure lives (since there might be no such surrounding space). It uses the fact that it is possible to define derivatives of real-valued functions on that geometric figure at a point p solely with the product rule and that the set of all such derivations in fact forms a vector space that is the desired tangent space.