Derivatives of Quotients
Example:
R(z) = (z + 1/4)2 / cos(z - 6)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| R(z) |
= |
(z + 1/4)2 |
 |
| cos(z - 6) |
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(z) |
= |
(z + 1/4)2 |
|
| cos(z - 6) |
we can think of as
| R(z) |
= |
f(z) |
|
| g(z) |
So the derivative is
| R(z)' |
= ( |
f(z) |
)' |
|
| g(z) |
| |
= |
f '(z) |
g(z) |
- |
f(z) |
g '(z) |
|
| ( g(z) )2 |
| |
= |
( (z + 1/4)2 )' |
( cos(z - 6) ) |
- |
( (z + 1/4)2 ) |
( cos(z - 6) )' |
|
| ( cos(z - 6) )2 |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( (z + 1/4)2 )' |
= |
2*(z + 1/4) (1 + 0) |
(by the chain rule) |
| ( cos(z - 6) )' |
= |
(-1)*sin(z - 6) (1 - 0) |
(by the chain rule) |
so the finished derivative is
| R(z)' |
= |
( 2*(z + 1/4) (1 + 0) ) |
( cos(z - 6) ) |
- |
( (z + 1/4)2 ) |
( (-1)*sin(z - 6) (1 - 0) ) |
|
| ( cos(z - 6) )2 |
| |
= |
2*(z + 1/4) cos(z - 6) + (z + 1/4)2 sin(z - 6) |
|
| (cos(z - 6))2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Wed Feb 4 08:57:38 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.