Derivatives of Quotients
Example:
R(p) = ln(p2 + 1) / (p + 1)5
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| R(p) |
= |
ln(p2 + 1) |
 |
| (p + 1)5 |
The Derivative Rule for Quotients:
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 our example,
| R(p) |
= |
ln(p2 + 1) |
|
| (p + 1)5 |
we can think of as
| R(p) |
= |
f(p) |
|
| g(p) |
So the derivative is
| R(p)' |
= ( |
f(p) |
)' |
|
| g(p) |
| |
= |
f '(p) |
g(p) |
- |
f(p) |
g '(p) |
|
| ( g(p) )2 |
| |
= |
( ln(p2 + 1) )' |
( (p + 1)5 ) |
- |
( ln(p2 + 1) ) |
( (p + 1)5 )' |
|
| ( (p + 1)5 )2 |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( ln(p2 + 1) )' |
= |
(p2 + 1)-1 (2*p + 0) |
(by the chain rule) |
| ( (p + 1)5 )' |
= |
5*(p + 1)4 (1 + 0) |
(by the chain rule) |
so the finished derivative is
| R(p)' |
= |
( (p2 + 1)-1 (2*p + 0) ) |
( (p + 1)5 ) |
- |
( ln(p2 + 1) ) |
( 5*(p + 1)4 (1 + 0) ) |
|
| ( (p + 1)5 )2 |
| |
= |
2*p (p2 + 1)-1 (p + 1)5 - 5*ln(p2 + 1) (p + 1)4 |
|
| ((p + 1)5)2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Sun Feb 1 01:29:14 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.