Derivatives of Quotients
Example:
f(t) = (sin(t) + sin(5)) / (10*t3 - 3*t2 - 1)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| f(t) |
= |
sin(t) + sin(5) |
 |
| 10*t3 - 3*t2 - 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,
| f(t) |
= |
sin(t) + sin(5) |
|
| 10*t3 - 3*t2 - 1 |
we can think of as
| f(t) |
= |
g(t) |
|
| h(t) |
So the derivative is
| f(t)' |
= ( |
g(t) |
)' |
|
| h(t) |
| |
= |
g '(t) |
h(t) |
- |
g(t) |
h '(t) |
|
| ( h(t) )2 |
| |
= |
( sin(t) + sin(5) )' |
( 10*t3 - 3*t2 - 1 ) |
- |
( sin(t) + sin(5) ) |
( 10*t3 - 3*t2 - 1 )' |
|
| ( 10*t3 - 3*t2 - 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
| f(t)' |
= |
( cos(t) + 0 ) |
( 10*t3 - 3*t2 - 1 ) |
- |
( sin(t) + sin(5) ) |
( 10*3*t2 - 3*2*t - 0 ) |
|
| ( 10*t3 - 3*t2 - 1 )2 |
| |
= |
cos(t) (10*t3 - 3*t2 - 1) - (sin(t) + sin(5)) (30*t2 - 6*t) |
|
| (10*t3 - 3*t2 - 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: Wed Feb 18 09:29:56 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.