Derivatives of Quotients
Example:
S(z) = 2z / (z2 + 1)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| S(z) |
= |
2z |
 |
| z2 + 1 |
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,
| S(z) |
= |
2z |
|
| z2 + 1 |
we can think of as
| S(z) |
= |
f(z) |
|
| g(z) |
So the derivative is
| S(z)' |
= ( |
f(z) |
)' |
|
| g(z) |
| |
= |
f '(z) |
g(z) |
- |
f(z) |
g '(z) |
|
| ( g(z) )2 |
| |
= |
( 2z )' |
( z2 + 1 ) |
- |
( 2z ) |
( z2 + 1 )' |
|
| ( z2 + 1 )2 |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
so the finished derivative is
| S(z)' |
= |
( (ln(2))*2z ) |
( z2 + 1 ) |
- |
( 2z ) |
( 2*z + 0 ) |
|
| ( z2 + 1 )2 |
| |
= |
(ln(2))*2z (z2 + 1) - 2*z 2z |
|
| (z2 + 1)2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Fri Mar 13 01:23:33 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.