Derivatives of Quotients
Example:
z = 2y / (sin(y) + 1)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| z |
= |
2y |
 |
| sin(y) + 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,
| z |
= |
2y |
 |
| sin(y) + 1 |
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 |
| |
= |
( 2y )' |
( sin(y) + 1 ) |
- |
( 2y ) |
( sin(y) + 1 )' |
 |
| ( sin(y) + 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
| z' |
= |
( (ln(2))*2y ) |
( sin(y) + 1 ) |
- |
( 2y ) |
( cos(y) + 0 ) |
 |
| ( sin(y) + 1 )2 |
| |
= |
(ln(2))*2y (sin(y) + 1) - 2y cos(y) |
 |
| (sin(y) + 1)2 |
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Wed Apr 8 10:09:08 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.