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In the previous section we noted that we had to be careful
when differentiating products or quotients.
It’s now time to look at products and quotients and see why.
First let’s take a look at why we have to be careful with
products and quotients. Suppose that we
have the two functions 
and . Let’s start by computing the derivative of
the product of these two functions. This
is easy enough to do directly.
Remember that on occasion we will drop the (x) part on the functions to simplify
notation somewhat. We’ve done that in
the work above.
Now, let’s try the following.
So, we can very quickly see that.
In other words, the derivative of a product is not the
product of the derivatives.
Using the same functions we can do the same thing for
quotients.
So, again we can see that,
To differentiate products and quotients we have the Product Rule and the Quotient Rule.
Product Rule
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If the two functions f(x)
and g(x) are differentiable (i.e. the derivative exist) then the
product is differentiable and,
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The proof of the Product Rule is shown in the Proof of Various Derivative Formulas section of
the Extras chapter.
Quotient Rule
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If the two functions
f(x) and g(x) are differentiable (i.e. the derivative exist) then the
quotient is differentiable and,
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Note that the numerator of the quotient rule is very similar
to the product rule so be careful to not mix the two up!
The proof of the Product Rule is shown in the Proof of Various Derivative Formulas section of
the Extras chapter.
Let’s do a couple of examples of the product rule.
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Example 1 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
Solution
At this point there really aren’t a lot of reasons to use
the product rule. As we noted in the
previous section all we would need to do for either of these is to just
multiply out the product and then differentiate.
With that said we will use the product rule on these so we
can see an example or two. As we add
more functions to our repertoire and as the functions become more complicated
the product rule will become more useful and in many cases required.
(a)
Note that we took the derivative of this function in the
previous section and didn’t
use the product rule at that point. We
should however get the same result here as we did then.
Now let’s do the problem here. There’s not really a lot to do here other
than use the product rule. However,
before doing that we should convert the radical to a fractional exponent as
always.
Now let’s take the derivative. So we take the derivative of the first
function times the second then add on to that the first function times the
derivative of the second function.
This is NOT what we got in the previous section for this
derivative. However, with some
simplification we can arrive at the same answer.
This is what we got for an answer in the previous section
so that is a good check of the product rule.
[Return to Problems]
(b)
This one is actually easier than the previous one. Let’s just run it through the product rule.
Since it was easy to do we went ahead and simplified the
results a little.
[Return to Problems]
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Let’s now work an example or two with the quotient
rule. In this case, unlike the product
rule examples, a couple of these functions will require the quotient rule in
order to get the derivative. The last
two however, we can avoid the quotient rule if we’d like to as we’ll see.
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Example 2 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
(a)
There isn’t a lot to do here other than to use the
quotient rule. Here is the work for
this function.
[Return to Problems]
(b)
Again, not much to do here other than use the quotient
rule. Don’t forget to convert the
square root into a fractional exponent.
[Return to Problems]
(c)
It seems strange to have this one here rather than being
the first part of this example given that it definitely appears to be easier
than any of the previous two. In fact,
it is easier. There is a point to
doing it here rather than first. In
this case there are two ways to do compute this derivative. There is an easy way and a hard way and in
this case the hard way is the quotient rule.
That’s the point of this example.
Let’s do the quotient rule and see what we get.
Now, that was the “hard” way. So, what was so hard about it? Well actually it wasn’t that hard, there is
just an easier way to do it that’s all.
However, having said that, a common mistake here is to do the
derivative of the numerator (a constant) incorrectly. For some reason many people will give the
derivative of the numerator in these kinds of problems as a 1 instead of 0! Also, there is some simplification that
needs to be done in these kinds of problems if you do the quotient rule.
The easy way is to do what we did in the previous section.
Either way will work, but I’d rather take the easier route
if I had the choice.
[Return to Problems]
(d)
This problem also seems a little out of place. However, it is here again to make a
point. Do not confuse this with a
quotient rule problem. While you can
do the quotient rule on this function there is no reason to use the quotient
rule on this. Simply rewrite the
function as
and differentiate as always.
[Return to Problems]
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Finally, let’s not forget about our applications of
derivatives.
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Example 3 Suppose
that the amount of air in a balloon at any time t is given by
Determine if the balloon is being filled with air or being
drained of air at .
Solution
If the balloon is being filled with air then the volume is
increasing and if it’s being drained of air then the volume will be
decreasing. In other words, we need to
get the derivative so that we can determine the rate of change of the volume
at .
This will require the quotient rule.
Note that we simplified the numerator more than usual
here. This was only done to make the
derivative easier to evaluate.
The rate of change of the volume at is then,
So, the rate of change of the volume at is negative and so the volume must be
decreasing. Therefore air is being
drained out of the balloon at .
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As a final topic let’s note that the product rule can be
extended to more than two functions, for instance.
With this section and the previous section we are now able
to differentiate powers of x as well
as sums, differences, products and quotients of these kinds of functions. However, there are many more functions out
there in the world that are not in this form.
The next few sections give many of these functions as well as give their
derivatives.