Derivatives of Quotients
Example:
Z = (y + 9)9 / sqrt(6*y - 1/4)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| Z |
= |
(y + 9)9 |
 |
| sqrt(6*y - 1/4) |
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,
| Z |
= |
(y + 9)9 |
|
| sqrt(6*y - 1/4) |
we can think of as
| Z |
= |
f(y) |
|
| g(y) |
So the derivative is
| Z' |
= ( |
f(y) |
)' |
|
| g(y) |
| |
= |
f '(y) |
g(y) |
- |
f(y) |
g '(y) |
|
| ( g(y) )2 |
| |
= |
( (y + 9)9 )' |
( sqrt(6*y - 1/4) ) |
- |
( (y + 9)9 ) |
( sqrt(6*y - 1/4) )' |
|
| ( sqrt(6*y - 1/4) )2 |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( (y + 9)9 )' |
= |
9*(y + 9)8 (1 + 0) |
(by the chain rule) |
| ( sqrt(6*y - 1/4) )' |
= |
(1/2)*(6*y - 1/4)-1/2 (6 - 0) |
(by the chain rule) |
so the finished derivative is
| Z' |
= |
( 9*(y + 9)8 (1 + 0) ) |
( sqrt(6*y - 1/4) ) |
- |
( (y + 9)9 ) |
( (1/2)*(6*y - 1/4)-1/2 (6 - 0) ) |
|
| ( sqrt(6*y - 1/4) )2 |
| |
= |
9*(y + 9)8 sqrt(6*y - 1/4) - 3*(y + 9)9 (6*y - 1/4)-1/2 |
|
| (sqrt(6*y - 1/4))2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
previous page
Page Generated: Wed Mar 11 02:36:35 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.