In the first section of this chapter we saw the definition of the derivative and we
computed a couple of derivatives using the definition. As we saw in those examples there was a fair
amount of work involved in computing the limits and the functions that we
worked with were not terribly complicated.
For more complex functions using the definition of the
derivative would be an almost impossible task.
Luckily for us we won’t have to use the definition terribly often. We will have to use it on occasion, however
we have a large collection of formulas and properties that we can use to
simplify our life considerably and will allow us to avoid using the definition
whenever possible.
We will introduce most of these formulas over the course of
the next several sections. We will start
in this section with some of the basic properties and formulas. We will give the properties and formulas in
this section in both “prime” notation and “fraction” notation.
Properties
Note that we have not included formulas for the derivative
of products or quotients of two functions here.
The derivative of a product or quotient of two functions is not the
product or quotient of the derivatives of the individual pieces. We will take a look at these in the next
section.
Next, let’s take a quick look at a couple of basic
“computation” formulas that will allow us to actually compute some derivatives.
Formulas
1)
If then OR
The derivative of a constant is
zero. See the Proof of Various Derivative
Formulas section of the Extras chapter to see the proof of this formula.
2)
If then OR ,
n is any number.
This formula is sometimes called
the power rule. All we are doing here is bringing the
original exponent down in front and multiplying and then subtracting one from
the original exponent.
Note as well that in order to use
this formula n must be a number, it
can’t be a variable. Also note that
the base, the x, must be a
variable, it can’t be a number. It
will be tempting in some later sections to misuse the Power Rule when we run
in some functions where the exponent isn’t a number and/or the base isn’t a
variable.
See the Proof of Various Derivative
Formulas section of the Extras chapter to see the proof of this formula. There are actually three different proofs
in this section. The first two
restrict the formula to n being an
integer because at this point that is all that we can do at this point. The third proof is for the general rule,
but does suppose that you’ve read most of this chapter.

These are the only properties and formulas that we’ll give
in this section. Let’s compute some
derivatives using these properties.
Example 1 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
(a)
In this case we have the sum and difference of four terms
and so we will differentiate each of the terms using the first property from
above and then put them back together with the proper sign. Also, for each term with a multiplicative
constant remember that all we need to do is “factor” the constant out (using
the second property) and then do the derivative.
Notice that in the third term the exponent was a one and
so upon subtracting 1 from the original exponent we get a new exponent of
zero. Now recall that . Don’t forget to do any basic arithmetic
that needs to be done such as any multiplication and/or division in the
coefficients.
[Return to Problems]
(b)
The point of this problem is to make sure that you deal
with negative exponents correctly.
Here is the derivative.
Make sure that you correctly deal with the exponents in
these cases, especially the negative exponents. It is an easy mistake to “go the other way”
when subtracting one off from a negative exponent and get instead of the correct .
[Return to Problems]
(c)
Now in this function the second term is not correctly set
up for us to use the power rule. The
power rule requires that the term be a variable to a power only and the term
must be in the numerator. So, prior to
differentiating we first need to rewrite the second term into a form that we
can deal with.
Note that we left the 3 in the denominator and only moved the
variable up to the numerator. Remember
that the only thing that gets an exponent is the term that is immediately to
the left of the exponent. If we’d
wanted the three to come up as well we’d have written,
so be careful
with this! It’s a very common mistake
to bring the 3 up into the numerator as well at this stage.
Now that we’ve gotten the function rewritten into a proper
form that allows us to use the Power Rule we can differentiate the
function. Here is the derivative for
this part.
[Return to Problems]
(d)
All of the terms in this function have roots in them. In order to use the power rule we need to
first convert all the roots to fractional exponents. Again, remember that the Power Rule
requires us to have a variable to a number and that it must be in the
numerator of the term. Here is the
function written in “proper” form.
In the last two terms we combined the exponents. You should always do this with this kind of
term. In a later section we will learn
of a technique that would allow us to differentiate this term without combining
exponents, however it will take significantly more work to do. Also don’t forget to move the term in the
denominator of the third term up to the numerator. We can now differentiate the function.
Make sure that you can deal with fractional
exponents. You will see a lot of them
in this class.
[Return to Problems]
(e)
In all of the previous examples the exponents have been
nice integers or fractions. That is
usually what we’ll see in this class.
However, the exponent only needs to be a number so don’t get excited
about problems like this one. They
work exactly the same.
The answer is a little messy and we won’t reduce the
exponents down to decimals. However,
this problem is not terribly difficult it just looks that way initially.
[Return to Problems]

There is a general rule about derivatives in this class that
you will need to get into the habit of using.
When you see radicals you should always first convert the radical to a
fractional exponent and then simplify exponents as much as possible. Following this rule will save you a lot of
grief in the future.
Back when we first put down the properties we noted that we
hadn’t included a property for products and quotients. That doesn’t mean that we can’t differentiate
any product or quotient at this point.
There are some that we can do.
Example 2 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
Solution
(a)
In this function we can’t just differentiate the first
term, differentiate the second term and then multiply the two back
together. That just won’t work. We will discuss this in detail in the next
section so if you’re not sure you believe that hold on for a bit and we’ll be
looking at that soon as well as showing you an example of why it won’t work.
It is still possible to do this derivative however. All that we need to do is convert the
radical to fractional exponents (as we should anyway) and then multiply this
through the parenthesis.
Now we can differentiate the function.
[Return to Problems]
(b)
As with the first part we can’t just differentiate the
numerator and the denominator and the put it back together as a
fraction. Again, if you’re not sure
you believe this hold on until the next section and we’ll take a more
detailed look at this.
We can simplify this rational expression however as
follows.
This is a function that we can differentiate.
[Return to Problems]

So, as we saw in this example there are a few products and
quotients that we can differentiate. If
we can first do some simplification the functions will sometimes simplify into
a form that can be differentiated using the properties and formulas in this
section.
Before moving on to the next section let’s work a couple of
examples to remind us once again of some of the interpretations of the
derivative.
Example 3 Is
increasing, decreasing or not changing at ?
Solution
We know that the rate of change of a function is given by
the functions derivative so all we need to do is it rewrite the function (to
deal with the second term) and then take the derivative.
Note that we
rewrote the last term in the derivative back as a fraction. This is not something we’ve done to this
point and is only being done here to help with the evaluation in the next
step. It’s often easier to do the
evaluation with positive exponents.
So, upon
evaluating the derivative we get
So, at the derivative is negative and so the
function is decreasing at .

Example 4 Find
the equation of the tangent line to at .
Solution
We know that the equation of a tangent line is given by,
So, we will need the derivative of the function (don’t
forget to get rid of the radical).
Again, notice that we eliminated the negative exponent in
the derivative solely for the sake of the evaluation. All we need to do then is evaluate the
function and the derivative at the point in question, .
The tangent line is then,

Example 5 The
position of an object at any time t
(in hours) is given by,
Determine when the object is moving to the right and when
the object is moving to the left.
Solution
The only way that we’ll know for sure which direction the
object is moving is to have the velocity in hand. Recall that if the velocity is positive the
object is moving off to the right and if the velocity is negative then the
object is moving to the left.
So, we need the derivative since the derivative is the
velocity of the object. The derivative
is,
The reason for factoring the derivative will be apparent
shortly.
Now, we need to determine where the derivative is positive
and where the derivative is negative.
There are several ways to do this.
The method that I tend to prefer is the following.
Since polynomials are continuous we know from the Intermediate Value Theorem that if the
polynomial ever changes sign then it must have first gone through zero. So, if we knew where the derivative was
zero we would know the only points where the derivative might change sign.
We can see from the factored form of the derivative that
the derivative will be zero at and . Let’s graph these points on a number line.
Now, we can see that these two points divide the number
line into three distinct regions. In each of these regions we know
that the derivative will be the same sign.
Recall the derivative can only change sign at the two points that are
used to divide the number line up into the regions.
Therefore, all that we need to do is to check the derivative
at a test point in each region and the derivative in that region will have
the same sign as the test point. Here
is the number line with the test points and results shown.
Here are the intervals in which the derivative is positive
and negative.
We included negative t’s
here because we could even though they may not make much sense for this
problem. Once we know this we also can
answer the question. The object is
moving to the right and left in the following intervals.

Make sure that you can do the kind of work that we just did
in this example. You will be asked
numerous times over the course of the next two chapters to determine where
functions are positive and/or negative.
If you need some review or want to practice these kinds of problems you
should check out the Solving
Inequalities section of my Algebra/Trig
Review.