Math Insight

A refresher on the product rule


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.


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