The Product Rule (Level 1) In the previous videos we learned that the
derivative of a sum and or difference of functions was equal to the sum and or difference of
the derivative of the individual functions. The next question we should ask is if this
pattern also works for the product of two functions. In other words is the derivative
of a product of two functions equal the product of the derivatives of the functions?
Let’s test this out with the following functions, let f(x) = x^3 and g(x) = x^6 the product
of these two functions is equal to x^9, and the derivative of this function is equal to
9x^8, now let’s verify the right side of this expression, so we go ahead and take the
derivative of f(x) and g(x) as follows, now if we were to multiply
them we should obtain 9x^8 but unfortunately we obtain something completely different in
this case 18x^7 this means that this expression does not hold. In other words the derivative
of a product is not the product of the derivatives. It turns out that in order to find the derivative
of functions formed by multiplication we have to make use of the product rule.
The product rule says that the derivative of a product of two functions is the first
function times the derivative of the second function plus the second function times the
derivative of the first function. In prime notation the product rule can be
written as: (fg)’ = fg’ + gf’. Remember the functions f and g have to be differentiable.
Also notice that it really does not matter which function you assign as your f(x) and
which one you assign as your g(x) as long as you are consistent with your choices everything
will work out. We can also rewrite the products rule as follows:
f’(x)g(x) + f(x)g’(x). Some students prefer this version because it generalizes easily
to products of three or more functions. For instance the product rule can be extended
to cover products involving more than two functions, for example if f, g, and h, are
differentiable functions of x, the derivative will be equal to: f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)
Notice that the derivative or prime symbol alternates between the functions this only
works if you keep the order of the functions consistent. Keep in mind that the derivative
of a product of two functions is not (in general) given by the product of the derivatives.
Alright with that said let’s see the product rule in action and go over some examples.
Find the derivative of: R(x) equals x cubed times e to the x.
The first thing you should notice is that we have a product of two functions.
In this case we have the power function x cubed and the exponential function. Since
we have a product of two functions we can find the derivative by making use of the product
rule. In order to apply the product rule we need to decide which function is going to be the
first function in this case f(x) and which one will be the second the function in this
case g(x). Keep in mind that it really does not matter which function you assign as your
f(x) and which function you assign as your g(x). It is totally up to you, just make sure
you apply the product rule correctly once you made your decision. For this example I
will assign x^3 as f(x) and e^x as g(x). With this assignment we go ahead and apply the
product rule. So we take f(x) our first function and multiply it by the derivative of g(x)
the second function then we add the product of g(x) and the derivative of f(x). Simplifying
the expression we obtain the final derivative equal to x^3 times e^x plus 3 times e^x times x^2
We can also rewrite this expression by factoring out an x^2 times e^x term. In any case both
forms are acceptable expressions for the derivative. Alright let’s try the next example.
Find the derivative of: R(x) equals e^x over the square root of x.
Notice that in this example we have a quotient of two functions, at first glance
it seems that we cannot use the product rule to find the derivative of this function. It
turns out that we can actually apply the product rule to this function all we need to do is
to rewrite the expression into a product rule friendly form. In order to accomplish this
we can rewrite the radical in the denominator as x raised to the power of negative 1/2, as
follows. Now we can go ahead and apply the product rule. I will assign e^x as my f(x)
the first function and assign x raised to the power of negative ½ as my g(x) the second
function, remember it really does not matter which function you assign as your f(x) and
g(x) as long as your consistent with the product rule everything will work out. Applying the
product rule we obtain the following expression, the first function times the derivative of
the second function plus the second function times the derivative of the fist function.
Simplifying and rewriting the expression we obtain the following for the derivative e^x
over the square root of x minus e^x over 2 times the square root of x cubed. Keep in
mind that you might need to rewrite an expression so that you can apply a specific differentiation
rule. Alright let’s try a slightly more challenging example.
Find the derivative of: R(x) equals e^x times the quantity x^2 minus 2x.
Here we have a product of two functions so let’s go ahead and use the product rule.
Let's assign e^x as our f(x) and assign the quantity x^2 - 2x as our g(x). Next we find
the derivative by applying the product rule so we take the first function and multiply
it by the derivative of the second function and add the second function and multiply it
by the derivative of the first function. Notice that in order to take the derivative of g(x)
we need to apply the sum rule, and power rule. Having applied the product rule we go ahead
and simplify the expression, here we can factor out an e^x term, this leaves us with the following,
then it is just a matter of simplifying the expression inside the brackets doing that
yields the final derivative equal to e^x times the quantity x^2 minus 2. Ok now let’s try
finding the derivative of a product of polynomials.
Find the derivative of: P(x) equals the quantity x + 3 times the quantity x^2 – 5x + 2
I hope you recognize this function this was the last function we worked on from the previous
video we can actually find the derivative of this function by using the product rule.
Let's go ahead and assign the binomial as our f(x) and the trinomial as our g(x) next we
go ahead and apply the product rule. So we take the first function and multiply it by
the derivative of the second function and add the second function times the derivative
of the first function. Next we go ahead and FOIL the expression on the left, then we go
ahead and simplify the expression by combining like terms. Doing that yields the final answer
equal to 3x^2 – 4x – 13. This is the same answer that we got in the previous video.
Notice that we could also find the derivative of this function by distributing the binomial
and then taking the derivative of each term like we did in the previous video, either
method will generate the same expression. Alright let’s try the final example.
Find the derivative of: P(x) equals the quantity x^3 – x times the quantity x^2 + 2 times the quantity
x^2 + x - 1. Notice that we have 3 functions being
multiplied. One way we can find the derivative is to apply the product rule by assigning
the product of two functions as our f(x) and assign the third function as our g(x) then
you would just carry out the product rule with these assignments. Taking this approach
would force you to apply an additional product rule when trying to find the derivative of
f(x) there is really nothing wrong with finding the derivative in this way. I actually encourage
you to find the derivative in this manner so you can practice applying the product rule
multiple times. We can also find the derivative by using the product rule extended for more
than two functions in this case we would need to assign 3 functions as our f(x), g(x) and
h(x), for this problem we will assign f(x) to be the first binomial, g(x) the second
binomial, and the trinomial as our h(x). Then it is just a matter of writing all the functions
making sure that we alternate the derivative to each distinct function, as follows, doing
that we obtain the following derivative of the expression. Technically we are done, this
crazy looking polynomial is our derivative, if you like you can go ahead and simplify
the expression by distributing and collecting like terms, but there is really no need to
do that, if for some reason you want simplify because you want to practice your distribution
and collecting like terms skills, the resulting polynomial you should obtain is as follows
7x^6 + 6x^5 + 4x^3 – 9x^2 - 4x + 2. Note that you can also just distribute the expressions
in the beginning and then apply the derivative term by term that would also yield the correct
answer. But for expressions that are not polynomials you want to use the product rule extended
for more than 2 functions. Alright and that is the product rule in a
nutshell. In our next video we will go over more challenging examples involving the product rule.