The Product Rule (Level 2) In this video we will go over more
challenging examples that require the use of the product rule. The following examples
require you to really understand the product rule and combine it with various algebra and
calculus concepts learned so far. The first type of problems involves writing an expression
for the derivative in terms of the given functions. Let’s take a look at the first example.
If p(x) and q(x) are differentiable functions, find an expression for the derivative of the
following function. R(x) = p(x)q(x). Notice that in this example we are given the
function R(x) which is written in terms of two other functions in this case p(x) and
q(x). The problem is asking us to find the derivative of this function. At first glance
it might be a little weird that we are asked to find the derivative of R(x) written in
this manner. It does not seem to contain any constants, power functions or natural exponential
functions. All we know is that the R(x) is defined by the product of p(x) and q(x). This
means that we need somehow work with these expressions and find the derivative. Since
R(x) is defined by a product of two functions it is natural to apply the product rule to
find the derivative of R(x). So we will assign p(x) as the first function and q(x) as the
second function. With these assignments we go ahead and apply the product rule, the derivative
of R(x) is going to be equal to p(x)q’(x) + q(x)p’(x)
and this is the final derivative, at first it might be hard to wrap your head around
this, if we knew what p(x) and q(x) was equal to we would still find the derivative
in this manner. All we are doing here is finding an expression for the derivative of R(x) which
was defined by two distinct functions p(x) and q(x). If we knew what p(x) and q(x) were
equal to then we would just substitute them into our expression. In this case we just
want a general expression for the derivative. Let’s try another example.
If p(x) and q(x) are differentiable functions, find an expression for the derivative of the
following function. R(x) = x^2q(x) + p(x). Alright notice that we now have a function
R(x) defined by a power function along with the functions q(x) and p(x). Similar to the
previous example we are asked to find an expression for the derivative. So we go ahead and take
the derivative of this function term by term. For the first term we have a product of two
functions in this case x squared and q(x). Since we have a product of two functions we
need to apply the product rule. We are going to assign x squared as the first function
and q(x) as the second function. Applying the product rule we obtain x squared times
q prime of x plus q(x) times 2x. Next we move along to the second term, the derivative of
the second term is just going to be equal to p prime of x. Rearranging the expressions
we obtain the final derivative equal to x^2q’(x) + 2xq(x) + p’(x)
Once again this expression represents the derivative of R(x) in terms of q(x) and p(x)
if we knew what P(x) and Q(x) were equal to we would just substitute them directly into
this expression. Alright let’s try one more example.
If p(x) and q(x) are differentiable functions, find an expression for the derivative of the
following function. R(x) = q(x)/x + e^xp(x) + 5x.
Similar to the previous examples let's go ahead and find the derivative of this expression.
This function is defined by 3 distinct terms; the first term contains the function q(x)
and the rational function 1 over x. In order to find the derivative of this term we need
to rewrite it into a derivative friendly form, in this case we can rewrite it as q(x) times
x raised to the power of negative 1. Now we can go ahead and apply the product rule with
q(x) as the first function and x raised to the power of negative 1 as the second function,
applying the product rule we obtain the following expression. Next we move along and find the
derivative of the second term. Here we also have a product of two functions in this case
the natural exponential function times p(x). So we go ahead and apply the product rule
with e^x as our first function and p(x) as our second function,
applying the product rule we obtain the following expression. Lastly we find the derivative
of 5x which is just equal to 5. Simplifying and rearranging the terms, we obtain the final
expression for the derivative which is equal to –q(x)/x^2 + q’(x)/x +e^xp’(x) + e^xp(x) + 5
Alright now let’s try a completely different type of problem.
Suppose g(2) = 3, g’(2) = -2, h(2) = -1, and h’(2) = 4, use this information to find
f’(2). Where f(x) = g(x)h(x) Notice that we are given values for the functions
g and h and their respective derivatives evaluated at x = 2. Also we are given that f(x) is equal
to the product of g(x) and h(x), this problem is asking us to find the derivative of f evaluated
at x = 2. Note that some text books might ask you this question by using the following
equivalent forms, the derivative of g times h evaluated at x = 2. Both expressions denote
the fact that we are looking for the derivative of g(x) times h(x) evaluated at x = 2. Alright,
at first this problem might seem confusing but the idea is to first find the derivative
of f(x) once we have an expression for the derivative in terms of g(x), g prime of x,
h(x) and h prime of x we go ahead and evaluate the derivative at x = 2 using the information
that is provided to us. So let’s go ahead and first find the derivative of f(x), since
this function is formed by a product of two other functions we make use of the product
rule, f prime of x is going to be equal to g(x) times h prime of x plus h(x) times g
prime of x. Having found the derivative we go ahead and evaluate the derivative at x
= 2 as follows. The next step is to replace the values for g(2) and g prime of 2 with 3 and
-2 respectively, and replace h prime of 2 and h(2) with 4 and -1 respectively. Then
it is just a matter of simplifying the expression, doing this we obtain the final answer equal
to 14. These types of problems requires you to first find the derivative of the function
in terms of the given functions then use the information that is provided to evaluate the
derivative at a given value of x.
Let's try another example. If f(x) = e^xh(x), where h(0) = 5 and h’(0)
= 2, find f’(0). Alright here we need to find the derivative
of f evaluated at x = 0, notice that this function is formed by the product of the natural
exponential function and h(x). Similar to the previous example let's first find an expression
for the derivative of f(x). Applying the product rule we obtain e^x times h prime of x plus
h(x) times e^x. Remember the derivative of the natural exponential function is just equal
to itself. Now let’s find the derivative of f evaluated at x = 0. Notice that we have
values for both h(0) and h prime of 0. So we go ahead and substitute those values into
our expression. We also want to evaluate the natural exponential function at x = 0. Doing
that we obtain the following, the final step is to simplify the numerical expression obtaining
7 as our final answer. These types of problems might seem a little
intimidating at first but as long as you understand that we are just looking for the derivative
in terms of other functions then the entire problem just becomes a standard find the derivative
problem, the only difference is that you are given values of the function at given values
of x. Alright in our next video we will go over
problems that require us to use graphs to find the derivative of a given function.