Derivatives of Quotients
Example:
G(s) = sqrt(10*s - 2) / sin(s)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| G(s) |
= |
sqrt(10*s - 2) |
 |
| sin(s) |
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,
| G(s) |
= |
sqrt(10*s - 2) |
|
| sin(s) |
we can think of as
| G(s) |
= |
f(s) |
|
| g(s) |
So the derivative is
| G(s)' |
= ( |
f(s) |
)' |
|
| g(s) |
| |
= |
f '(s) |
g(s) |
- |
f(s) |
g '(s) |
|
| ( g(s) )2 |
| |
= |
( sqrt(10*s - 2) )' |
( sin(s) ) |
- |
( sqrt(10*s - 2) ) |
( sin(s) )' |
|
| ( sin(s) )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
| G(s)' |
= |
( (1/2)*(10*s - 2)-1/2 (10 - 0) ) |
( sin(s) ) |
- |
( sqrt(10*s - 2) ) |
( cos(s) ) |
|
| ( sin(s) )2 |
| |
= |
5*(10*s - 2)-1/2 sin(s) - sqrt(10*s - 2) cos(s) |
|
| (sin(s))2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Tue Jan 27 07:10:49 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.