(d/dx)( f(x)g(x) ) = f(x)( (d/dx)g(x) ) + g(x)( (d/dx)f(x) )

What this means is that the derivative of a product equals the first factor times the derivative of the second factor, plus the second times the derivative of the first.

Proof:

Letting y = f(x)g(x) and using the definition of a derivative:

    f(x+h) - f(x)
lim -------------
h→0     h

we write

dy        f(x+h)g(x+h) - f(x)g(x)
-- = lim ------------------------.
dx   h→0            h

To change the form of the quotient so that the limit may be evaluated, we subtract and 
add the expression f(x+h)g(x) in the numerator.

dy       f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)
-- = lim ------------------------------------------------- =
dx   h→0                        h

         _                                               _
        |          g(x+h) - g(x)           f(x+h) - f(x) |     
     lim|f(x+h) * -------------- + g(x) * -------------- |=
        |h→0           h                       h         |
        |_                                              _|

                      g(x+h) - g(x)                  f(x+h) - f(x)
     lim f(x+h) * lim ------------- + lim g(x) * lim ------------
     h→0                   h                              h

Since f is differentiable at x, it is continuous at x. Hence, lim_h→0 f(x+h) = f(x). Also, lim_h→0 g(x) = g(x), since x is fixed in this limiting process. Finally, applying the definition of derivative to f(x) and g(x), we obtain

dy
-- = f(x)g'(x) + g(x)f'(x).
dx

The product rule is good to use if the factors of the product are too complicated to multiply out beforehand. However, we'll start with an easy exercise, for other rules might have to be known, such as the chain rule, for a complicated exercise.

Example:

y = (x³ + 1)(2x² + 8x - 5)

(dy/dx)
= (x³ + 1)(dy/dx)(2x² + 8x - 5) + (2x² + 8x - 5)(dy/dx)(x³ + 1)

= (x³ + 1)(4x + 8) + (2x² + 8x - 5)(3x²)

= (4x⁴ + 8x³ + 4x + 8) + (6x⁴ + 24x³ - 15x²)

= 10x⁴ + 32x³ - 15x² + 4x + 8

This answer could also be found if y was multiplied out and then derived, which is a good way to check the answers of these exercises.

When studying multivariable calculus, one encounters a whole zoo of product rules: there's one for dot products, there's one for cross products, there's one for matrix products; later in functional analysis one sees yet another product rule for inner products. After a while, one gets the feeling that pretty much every notion of product comes with its own product rule. Isn't there some underlying general rule? Yes there is:

Let X, Y and Z be Banach spaces and let B : X × Y → Z be a continuous bilinear operator. Then B is differentiable and the derivative at the point (x,y) ∈ X×Y is the linear map D_(x,y)B : X × Y → Z defined by D_(x,y)B(u,v) = B(x,v) + B(u,y) for every (u,v)∈X×Y.

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This writeup is in the public domain.

The following product rules are used when calculating derivatives of scalar functions and vector fields in multiple dimensions. They can be shown by breaking both sides of the equalities into components and using the product rule for single variables (as explained above): Let g,h be differentiable scalar functions of x. Then d/dx(gh)=g(dh/dx)+h(dg/dx).

A few words about notation:

• All vector fields are in bold, all scalar functions are not.
• gh is the product of g and h, and gA is the product of g and A.
• A·B is the dot product of A with B.
• A×B is the cross product of A with B.
• ∇g is the gradient of g.
• ∇·A is the divergence of A.
• ∇×A is the curl of A.
• (A·∇)B is the vector [A_x1∂B_x1/∂x1, A_x2∂B_x2/∂x2, ...]

It makes sense that we would need 6 product rules for multivariable differentiation. We have three operators: gradient, divergence and curl. Gradient operates on a scalar function while divergence and curl operate on vector fields. A scalar can be formed by the product of two scalars functions or by the dot product of two vector fields. A vector can be formed from the product of a scalar function and a vector field or by the cross product of two vector fields. Therefore, there are three operations, each of which can operate on one of two possible combinations of scalar functions and vector fields, and three times two is six.

Let g,h be differentiable scalar functions of x₁,x₂,...,x_n, and let A,B be differentiable vector fields in Rⁿ.

1. ∇(gh)=g∇h+h∇g
2. ∇(A·B)=A×(∇×B)+B×(∇×A)+(A·∇)B+(B·∇)A
3. ∇·(gA)=g(∇·A)+A·(∇g)
4. ∇·(A×B)=B·(∇×A)-A·(∇×B)
5. ∇×(gA)=g(∇×A)+A×(∇g)
6. ∇×(A×B)=(B·∇)A-(A·∇)B+A(∇·B)-B(∇·A)

Sources:
Griffiths, D. Introduction to Electrodynamics Third Edition © 1999.
http://www.pas.rochester.edu/~orr/p217/lecture2.pdf

An alternative proof of the product rule

The standard proof of the product rule that you'll find in every calculus textbook under the sun uses the curious trick of adding and subtracting the same term to facilitate factoring. It is historically accurate (the way Leibniz did it) and simple. As cute as this trick is (and how typical of analysis!), it would be nice to have a more intuitive proof of the product rule that didn't rely on such... deviousness.

This proof of the product rule uses Caratheodory's lemma to deal with the unpleasantries that the aforementioned trick handled. As a reminder, Caratheodory's lemma says that any function f(x) differentiable at a point c can be re-written as f(c) + φ(x)*(x-c), and φ will be a function continuous at c.

Let us restate the product rule again, in a somewhat different form:

If f and g are differentiable at the same point c,
then (f*g)'(c) exists and is f'(c)*g(c) + f(c)*g'(c).

The proof follows from manipulating the definition of the derivative of (f*g):

            f(x)g(x) - f(c)g(c)
f'(c) = lim -------------------
        x→c        x - c

Now this is the part (as you can see elsewhere in this node) where you add f(c)g(x) - f(c)g(x) to the numerator to factor things. Instead, we apply the lemma, using φ for the function corresponding to f and ψ for the function corresponding to g in the lemma.

            (f(c) + φ(x)*(x-c))(g(c) + ψ(x)*(x-c)) - f(c)g(c)
f'(c) = lim -------------------------------------------------
        x→c                      x - c

Now foil out the first term in the numerator:

            f(c)g(c) + f(c)ψ(x)(x-c) + g(c)φ(x)(x-c) + φ(x)ψ(x)(x-c)(x-c) - f(c)g(c)
f'(c) = lim -------------------------------------------------------------------------
        x→c                               x - c

The first and last terms in the numerator cancel, and we can proceed to carry out the division.

f'(c) = lim f(c)ψ(x) + g(c)φ(x) + φ(x)ψ(x)(x-c)
        x→c

Remember φ and ψ are continuous at c, so we can take the limit:

f'(c) = f(c)ψ(c) + g(c)φ(c)

And the lemma provides that, at c, both φ and ψ agree with their corresponding function's derivatives.

f'(c) = f(c)g'(c) + f'(c)g(c)

Much the same method can be used to prove the quotient rule and the chain rule, but I've found few high school math teachers who are familiar with Caratheodory's lemma, so using this proof on a calculus final is somewhat harder to justify. It is still interesting to see different proofs of the same theorem, from an aesthetic point of view.

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Thanks to my analysis prof, who taught me the lemma despite it being "a useless bit of trivium."