Derivatives of Quotients
When finding any derivative, we have to first recognize the
pattern that tells us what rule to use, and then apply the
rule to find the derivative. For quotients, seeing the pattern
(that there is a quotient) is rather easier than it is for some other
rules. Once we see that, we just need to apply the rule.
a few popular functions
| |
power |
: |
x r |
| |
polynomial |
: |
c n
x n +
c n-1
x n-1 + ... +
c 0 |
| |
exponential |
: |
a x |
| |
natural log |
: |
ln(x) |
| |
sine |
: |
sin(x) |
| |
cosine |
: |
cos(x) |
| |
tangent |
: |
tan(x) |
Let's consider some examples. A quotient is just one function divided
by another. Some of the functions we work with most frequently are
shown in the box to the right, and examples of quotients involving
these are
sin(x) / e2x,
(1 + x) / (3 - x), and
5 / ln(x).
Is it obvious that the last of these is a quotient? It looks like it
written this way, but if we rewrite it as 5(ln(x))-1, it also looks like a
constant multiple,
with 5 multiplying a
chain rule
problem. In practice, that's probably an easier way to do this
problem, but for now we'll do it with the quotient rule. The answer
that we get will look a little different, but after simplification are
the same thing.
Applying the Rule:
This is the quotient rule:
The derivative of a quotient is the derivative of the numerator
times the denominator minus the numerator times the derivative of the
denominator, all divided by the denominator squared.
If
| |
z |
= ( |
f(x) |
) |
 |
| g(x) |
then the derivative of
z is
| |
z ' |
= ( |
f(x) |
)' |
|
| g(x) |
| |
|
= |
f '(x) g(x) |
- |
f(x) g '(x) |
|
|
|
( g(x) )2 |
So to apply the quotient rule, we need to
- find the derivatives of the numerator and denominator,
and
- plug them into the quotient rule.
Let's do this for the examples above. For step one, let's make a
table of the numerators, denominators, and their derivatives:
| f(x) / g(x) |
|
|
|
numerator (f) |
|
|
|
denominator (g) |
|
|
|
f '(x) |
|
|
|
g '(x) |
|
| sin(x) / ex |
|
|
|
sin(x) |
|
|
|
e2x |
|
|
|
cos(x) |
|
|
|
2 e2x |
| (1 + x) / (3 - x) |
|
|
|
1 + x |
|
|
|
3 - x |
|
|
|
1 |
|
|
|
-1 |
| 5 / ln(x |
|
|
|
5 |
|
|
|
ln(x) |
|
|
|
0 |
|
|
|
1 / x |
Then we just have to plug things into the quotient rule. Remember
the minus sign! Forgetting it has caused pain and grief for many
a calculus student. The quotient rule is:
| ( |
f(x) |
) ' |
= |
f '(x) |
g(x) |
- |
f(x) |
g '(x) |
|
|
| g(x) |
|
(g(x)) 2
|
|
So, plugging in from our table of numerators, denominators and
derivatives, we get
| ( |
sin(x) |
)' |
= |
cos(x) |
e2x |
- |
sin(x) |
2 e2x |
|
|
| e2x |
|
(e2x)2 |
|
|
|
| ( |
1 + x |
)' |
= |
1 |
(3 - x) |
- |
(1 + x) |
(-1) |
|
|
| 3 - x |
|
(3 - x)2 |
|
|
| ( |
5 |
)' |
= |
0 |
ln(x) |
- |
5 |
(1 / x) |
= |
-5 |
|
|
|
| ln(x) |
|
(ln(x))2 |
|
x
(ln(x))2 |
|
Summary:
Let's summarize the steps we take to find derivatives of
quotients.
- Recognize the quotient, and find the numerator and
denominator
- Find the derivatives we need: then we find the
derivatives of each of these functions.
- Assemble the derivative: and then we use the quotient
rule to assemble the derivative. And we remember the negative
sign!
![[]](images/boxby.gif)
see a quotient rule example
practice test on the quotient rule
practice gateway test
previous
page
Deriv Tutorials: Quotients
Last Modified: Mon May 7 13:27:46 EDT 2001
Comments to
glarose@umich.edu
©2001 Gavin LaRose, UM Math Dept.