Product rule

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In mathematics, the product rule of calculus, which is also called Leibniz's law (see derivation), governs the differentiation of products of differentiable functions.

It may be stated thus:

or in the Leibniz notation thus:

Contents

1 Discovery by Leibniz

2 Examples

3 Common error

4 Proof of the product rule

5 Generalizations

6 See also

Discovery by Leibniz

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

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

which can also be written in "prime notation" as

Examples

Suppose you want to differentiate f(x) = x² sin(x). By using the product rule, you get the derivative f'(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 f(x) is a differentiable function, then cf(x) is also differentiable, and its derivative is (c × f)'(x) = c × f'(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 product rule can be used to derive the rule for integration by parts and the quotient rule.

Common error

It is a common error, when studying calculus, to suppose that the derivative of (uv) equals (u′)(v′) (Leibniz himself made this error initially); 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.

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 quotients:

Suppose

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

<math>f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{g(x + \Delta x)h(x + \Delta x) - g(x)h(x)}{\Delta x}<math>

Since

<math>[g(x)h(x + \Delta x) - g(x)h(x + \Delta x)]/(\Delta x) = 0<math>,

we have

<math>f'(x) = \lim_{\Delta x \to 0} \frac{g(x + \Delta x)h(x + \Delta x) - g(x)h(x) + g(x)h(x + \Delta x) - g(x)h(x + \Delta x)}{\Delta x}<math>

<math>= \lim_{\Delta x \to 0} \frac{g(x)(h(x + \Delta x) - h(x)) + h(x + \Delta x)(g(x + \Delta x) - g(x))}{\Delta x}<math>

<math>= \lim_{\Delta x \to 0} \left[g(x)\left(\frac{(h(x + \Delta x) - h(x))}{\Delta x}\right) + h(x + \Delta x)\left(\frac{(g(x + \Delta x) - g(x))}{\Delta x}\right)\right]<math>

Since h is continuous at x, we have

<math>\lim_{\Delta x \to 0} h(x + \Delta x) = h(x)<math>

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

<math>\left[\lim_{\Delta x \to 0} \frac{(h(x + \Delta x) - h(x))}{\Delta x}\right] = h'(x) \mbox{ and } \left[\lim_{\Delta x \to 0} \frac{(g(x + \Delta x) - g(x))}{\Delta x}\right] = g'(x)<math>

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

<math>f'(x) = \lim_{\Delta x \to 0} \left[g(x)\left(\frac{(h(x + \Delta x) - h(x))}{\Delta x}\right) + h(x + \Delta x)\left(\frac{(g(x + \Delta x) - g(x))}{\Delta x}\right)\right]<math>

<math>= \left[\lim_{\Delta x \to 0} g(x)\right]\left[\lim_{\Delta x \to 0} \frac{(h(x + \Delta x) - h(x))}{\Delta x}\right] + \left[\lim_{\Delta x \to 0} h(x + \Delta x)\right]\left[\lim_{\Delta x \to 0}\frac{(g(x + \Delta x) - g(x))}{\Delta x}\right]<math>

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

For a collection of functions <math>f_1 \dots f_k<math>, we can write this more succinctly as

<math>\frac{d}{dx} \prod_{i=1}^k f_i(x)

= \left(\sum_{i=1}^k \frac{\frac{d}{dx} f_i(x)}{f_i(x)}\right)
  \prod_{i=1}^k f_i(x)<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, often called the Leibniz rule, 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

<math> D_\left( x,y \right)\,B_\left( u,v \right) = B_\left( x,y \right) + B_\left( u,v \right)\;\mbox{for all}\;(u,v)\;\mbox{in}\;X \times Y <math>.