# Product rule

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:

[itex]\,\!(fg)'=f'g+fg'[itex]

or in the Leibniz notation thus:

[itex]{d\over dx}(uv)=u{dv\over dx}+v{du\over dx}.[itex]
 Contents

## 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

[itex]d(uv) = (u + du)(v + dv) - uv = u(dv) + v(du) + (du)(dv)[itex]

Since the term (du)(dv) is "negligible" (i.e. at least quadratic in du and dv), Leibniz concluded that

[itex]d(uv) = (du)v + u(dv)[itex]

and this is indeed the differential form of the product rule. If we divide through by the differential dx, we obtain

[itex]\frac{d}{dx} (uv) = \left( \frac{du}{dx} \right) v + u \left( \frac{dv}{dx} \right)[itex]

which can also be written in "prime notation" as

[itex](uv)' = u' v + u v'[itex]

This can also be seen as a 'barber shop' analogy. For example, in the above example, [itex]u[itex] stands at one side while [itex]v[itex] takes the 'haircut'.

## 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

[itex]f(x) = g(x)h(x)[itex]

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

[itex]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}[itex]

Since

[itex]g(x + \Delta x)h(x + \Delta x) - g(x)h(x) = g(x)(h(x + \Delta x) - h(x)) + h(x + \Delta x)(g(x + \Delta x) - g(x)),

[itex]

we have

[itex]f'(x) = \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}[itex]
[itex]= \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][itex]

Since h is continuous at x, we have

[itex]\lim_{\Delta x \to 0} h(x + \Delta x) = h(x)[itex]

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

[itex]\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)[itex]

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

[itex]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][itex]
[itex]= \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][itex]
[itex]= g(x)h'(x) + h(x)g'(x)[itex]

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

[itex]\frac{d(uvw)}{dx} = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}[itex]

For a collection of functions [itex]f_1 \dots f_k[itex], we can write this more succinctly as

[itex]\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)[itex]


It can also be generalized to Leibniz rule for higher derivatives of a product of two factors: if y = uv and y(n) denotes the n-th derivative of y, then

[itex]y^{(n)}(x) = \sum_{k=0}^n {n \choose k} u^{(n-k)}(x)\; v^{(k)}(x)[itex] ,

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 × YZ 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 × YZ given by

[itex] 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 [itex].

### Derivation in abstract algebra

In abstract algebra, the product rule is used to define what is called a derivation, not vice versa. A particularly important (and still very abstract) example are Lie algebras, extensively used in modern theoretical physics, especially gauge field theory.

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy