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With this section we’re going to start looking at the
derivatives of functions other than polynomials or roots of polynomials. We’ll start this process off by taking a look
at the derivatives of the six trig functions.
Two of the derivatives will be derived.
The remaining four are left to the reader and will follow similar proofs
for the two given here.
Before we actually get into the derivatives of the trig
functions we need to give a couple of limits that will show up in the
derivation of two of the derivatives.
Fact
See the Proof of Trig Limits
section of the Extras chapter to see the proof of these two limits.
Before we start differentiating trig functions let’s work a
quick set of limit problems that this fact now allows us to do.
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Example 1 Evaluate
each of the following limits.
(a)  [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
(f) [Solution]
Solution
(a)
There really isn’t a whole lot to this limit. In fact, it’s only here to contrast with
the next example so you can see the difference in how these work. In this case since there is only a 6 in the
denominator we’ll just factor this out and then use the fact.
[Return to Problems]
(b)
Now, in this case we can’t factor the 6 out of the sine so
we’re stuck with it there and we’ll need to figure out a way to deal with
it. To do this problem we need to
notice that in the fact the argument of the sine is the same as the
denominator (i.e. both ’s).
So we need to get both of the argument of the sine and the denominator
to be the same. We can do this by
multiplying the numerator and the denominator by 6 as follows.
Note that we
factored the 6 in the numerator out of the limit. At this point, while it may not look like
it, we can use the fact above to finish the limit.
To see that we
can use the fact on this limit let’s do a change of variables. A
change of variables is really just a renaming of portions of the problem to
make something look more like something we know how to deal with. They can’t always be done, but sometimes,
such as this case, they can simplify the problem. The change of variables here is to let and then notice that as we also have
. When doing a change of variables in a limit
we need to change all the x’s into ’s and that includes the one in the
limit.
Doing the
change of variables on this limit gives,
And there we are.
Note that we didn’t really need to do a change of variables here. All we really need to notice is that the
argument of the sine is the same as the denominator and then we can use the
fact. A change of variables, in this
case, is really only needed to make it clear that the fact does work.
[Return to Problems]
(c)
In this case we appear to have a small problem in that the
function we’re taking the limit of here is upside down compared to that in
the fact. This is not the problem it
appears to be once we notice that,
and then all we
need to do is recall a nice property of limits that allows us to do ,
With a little
rewriting we can see that we do in fact end up needing to do a limit like the
one we did in the previous part. So,
let’s do the limit here and this time we won’t bother with a change of
variable to help us out. All we need
to do is multiply the numerator and denominator of the fraction in the
denominator by 7 to get things set up to use the fact. Here is the work for this limit.
[Return to Problems]
(d)
This limit looks nothing like the limit in the fact,
however it can be thought of as a combination of the previous two parts by
doing a little rewriting. First, we’ll
split the fraction up as follows,
Now, the fact
wants a t in the denominator of the
first and in the numerator of the second.
This is easy enough to do if we multiply the whole thing by (which is just one after all and so won’t
change the problem) and then do a little rearranging as follows,
At this point
we can see that this really is two limits that we’ve seen before. Here is the work for each of these and
notice on the second limit that we’re going to work it a little differently
than we did in the previous part. This
time we’re going to notice that it doesn’t really matter whether the sine is
in the numerator or the denominator as long as the argument of the sine is
the same as what’s in the numerator the limit is still one.
Here is the
work for this limit.
[Return to Problems]
(e)
This limit almost looks the same as that in the fact in
the sense that the argument of the sine is the same as what is in the
denominator. However, notice that, in
the limit, x is going to 4 and not
0 as the fact requires. However, with
a change of variables we can see that this limit is in fact set to use the
fact above regardless.
So, let and then notice that as we have . Therefore, after doing the change of
variable the limit becomes,
[Return to Problems]
(f)
The previous parts of this example all used the sine
portion of the fact. However, we could
just have easily used the cosine portion so here is a quick example using the
cosine portion to illustrate this.
We’ll not put in much explanation here as this really does work in the
same manner as the sine portion.
All that is
required to use the fact is that the argument of the cosine is the same as
the denominator.
[Return to Problems]
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Okay, now that we’ve gotten this set of limit examples out
of the way let’s get back to the main point of this section, differentiating
trig functions.
We’ll start with finding the derivative of the sine
function. To do this we will need to use
the definition of the derivative. It’s
been a while since we’ve had to use this, but sometimes there just isn’t
anything we can do about it. Here is the
definition of the derivative for the sine function.
Since we can’t just
plug in to evaluate the limit we will need to
use the following trig formula on the first sine in the numerator.
Doing this gives us,
As you can see upon using the trig formula we can combine
the first and third term and then factor a sine out of that. We can then break up the fraction into two
pieces, both of which can be dealt with separately.
Now, both of the limits here are limits as h approaches zero. In the first limit we have a sin(x) and in the second limit we have a
cos(x). Both of these are only functions of x only and as h moves in towards zero this has no affect on the value of x.
Therefore, as far as the limits are concerned, these two functions are
constants and can be factored out of their respective limits. Doing this gives,
At this point all we need to do is use the limits in the
fact above to finish out this problem.
Differentiating cosine is done in a similar fashion. It will require a different trig formula, but
other than that is an almost identical proof.
The details will be left to you.
When done with the proof you should get,
With these two out of the way the remaining four are fairly
simple to get. All the remaining four
trig functions can be defined in terms of sine and cosine and these
definitions, along with appropriate derivative rules, can be used to get their
derivatives.
Let’s take a look at tangent. Tangent is defined as,
Now that we have the derivatives of sine and cosine all that
we need to do is use the quotient rule on this.
Let’s do that.
Now, recall that and if we also recall the definition of secant
in terms of cosine we arrive at,
The remaining three trig functions are also quotients
involving sine and/or cosine and so can be differentiated in a similar
manner. We’ll leave the details to
you. Here are the derivatives of all six
of the trig functions.
Derivatives of the
six trig functions
At this point we should work some examples.
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Example 2 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
There really isn’t a whole lot to this problem. We’ll just differentiate each term using
the formulas from above.
[Return to Problems]
(b)
In this part we will need to use the product rule on the
second term and note that we really will need the product rule here. There is no other way to do this derivative
unlike what we saw when we first looked at the product rule. When we first looked at the product rule
the only functions we knew how to differentiate were polynomials and in those
cases all we really needed to do was multiply them out and we could take the
derivative without the product rule.
We are now getting into the point where we will be forced to do the
product rule at times regardless of whether or not we want to.
We will also need to be careful with the minus sign in
front of the second term and make sure that it gets dealt with properly. There are two ways to deal with this. One way it to make sure that you use a set
of parenthesis as follows,
Because the
second term is being subtracted off of the first term then the whole
derivative of the second term must also be subtracted off of the derivative
of the first term. The parenthesis
make this idea clear.
A potentially easier way to do this is to think of the
minus sign as part of the first function in the product. Or, in other words the two functions in the
product, using this idea, are and . Doing this gives,
So, regardless of how you approach this problem you will
get the same derivative.
[Return to Problems]
(c)
As with the previous part we’ll need to use the product
rule on the first term. We will also
think of the 5 as part of the first function in the product to make sure we
deal with it correctly. Alternatively,
you could make use of a set of parenthesis to make sure the 5 gets dealt with
properly. Either way will work, but
we’ll stick with thinking of the 5 as part of the first term in the
product. Here’s the derivative of this
function.
[Return to Problems]
(d)
In this part we’ll need to use the quotient rule to take
the derivative.
Be careful with the signs when differentiating the
denominator. The negative sign we get
from differentiating the cosine will cancel against the negative sign that is
already there.
This appears to be done, but there is actually a fair
amount of simplification that can yet be done. To do this we need to factor out a “-2”
from the last two terms in the numerator and the make use of the fact that .
[Return to Problems]
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As a final problem here let’s not forget that we still have
our standard interpretations to derivatives.
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Example 3 Suppose
that the amount of money in a bank account is given by
where t is in
years. During the first 10 years in
which the account is open when is the amount of money in the account
increasing?
Solution
To determine when the amount of money is increasing we
need to determine when the rate of change is positive. Since we know that the rate of change is
given by the derivative that is the first thing that we need to find.
Now, we need to determine where in the first 10 years this
will be positive. This is equivalent
to asking where in the interval [0, 10] is the derivative positive. Recall that both sine and cosine are
continuous functions and so the derivative is also a continuous
function. The Intermediate Value Theorem then tells us that
the derivative can only change sign if it first goes through zero.
So, we need to solve the following equation.
The solution to this equation is,
If you don’t recall how to solve trig equations go back
and take a look at the sections on solving
trig equations in the Review chapter.
We are only interested in those solutions that fall in the
range [0, 10]. Plugging in values of n into the solutions above we see that
the values we need are,
So, much like solving polynomial inequalities all that we
need to do is sketch in a number line and add in these points. These points will divide the number line
into regions in which the derivative must always be the same sign. All that we need to do then is choose a
test point from each region to determine the sign of the derivative in that
region.
Here is the number line with all the information on it.
So, it looks like the amount of money in the bank account
will be increasing during the following intervals.
Note that we can’t say anything about what is happening
after since we haven’t done any work for t’s after that point.
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In this section we saw how to differentiate trig
functions. We also saw in the last
example that our interpretations of the derivative are still valid so we can’t
forget those.
Also, it is important that we be able to solve trig
equations as this is something that will arise off and on in this course. It is also important that we can do the kinds
of number lines that we used in the last example to determine where a function
is positive and where a function is negative.
This is something that we will be doing on occasion in both this chapter
and the next.