MATH 135 – Calculus 1 The Derivative Product Rule October 21, 2019

Background

In today’s video, we saw that the derivative of a product of functions is given by

(f · g)′(x) = f(x)g′(x) + g(x)f′(x),

provided that f and g are each differentiable at x. (In other words, if f, g are differentiable separately, then so is the product, and the derivative of f · g is given by the formula above.) Today, we want to practice with this new derivative rule

Questions

(1) Differentiate by the product rule:

(a) (x^{2} + 3x + 1)(x^{3} + x + 2). Then, expand out the product, differentiate using our more basic rules, and compare your answers. They should be equal.

(b) (x^{2} + x + 1)e^{x}.

(c) e^{2x} (How can you break this up as an f(x)g(x)?)

(d) (x^{1/2} + 3x)(x − 4e^{x}).

(2) Differentiate using the product rule “in two stages:”

(x^{2} + 3x)(e^{x} + x)(x + x^{−2}).

(3) Now suppose you have a product y = f(x)g(x)h(x) with three factors in general. If you group

the terms like this:

f(x)g(x)h(x) = f(x) · (g(x)h(x)),

explain why

y′ = f(x)(g(x)h(x))′ + (g(x)h(x))f′(x).

Now work out (g(x)h(x))′ by the product rule and rewrite things to see

y′ = f(x)g(x)h′(x) + f(x)g′(x)h(x) + f′(x)g(x)h(x).

(4) Continuing from (3), what is the pattern here? What if you have a product with any number of factors y = f_{1}(x)f_{2}(x) ⋯ f_{n}(x). What will the derivative of that product look like?