Not only will the *product rule* be of use in general and later
on, but it's already helpful in perhaps unexpected ways in dealing
with polynomials. Anyway, the general rule is
$${d\over dx}(fg)=f'g+fg'$$
While this is certainly not as awful as the quotient rule,
it is not as simple as the rule for sums, which was the good-sounding
slogan that *the derivative of the sum is the sum of the
derivatives*. It is *not* true that the derivative of the product
is the product of the derivatives. Too bad. Still, it's not as bad as
the quotient rule.

One way that the product rule can be useful is in postponing or
eliminating a lot of algebra. For example, to evaluate
$${d\over dx}\left((x^3+x^2+x+1)(x^4+x^3+2x+1)\right)$$
we *could* multiply out and then take the derivative term-by-term
as we did with several polynomials above. This would be at least
mildly irritating because we'd have to do a bit of algebra. Rather,
just apply the product rule *without* feeling compelled first to
do any algebra:
$${d\over dx}\left((x^3+x^2+x+1)(x^4+x^3+2x+1)\right)$$
$$=(x^3+x^2+x+1)'(x^4+x^3+2x+1)+(x^3+x^2+x+1)(x^4+x^3+2x+1)'$$
$$=(3x^2+2x+1)(x^4+x^3+2x+1)+(x^3+x^2+x+1)(4x^3+3x^2+2)$$
Now if we were somehow still obliged to multiply out, then we'd still
have to do some algebra. But *we can take the derivative without
multiplying out*, if we want to, by using the product rule.

For that matter, once we see that there is a *choice* about doing
algebra either *before* or *after* we take the derivative, it
might be possible to make a choice which minimizes our computational
labor. This could matter.

#### Exercises

- Find ${ d \over dx }(x^3-1)(x^6+x^3+1))$
- Find ${ d \over dx }(x^2+x+1)(x^4-x^2+1)$.
- Find ${ d \over dx }(x^3+x^2+x+1)(x^4+x^2+1))$
- Find ${ d \over dx }(x^3+x^2+x+1)(2x + \sqrt{x}))$