Derivatives of Quotients
Example:
B = ex + p / sin((p)*x)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| B |
= |
ex + p |
 |
| sin((p)*x) |
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,
| B |
= |
ex + p |
|
| sin((p)*x) |
we can think of as
| B |
= |
f(x) |
|
| g(x) |
So the derivative is
| B' |
= ( |
f(x) |
)' |
|
| g(x) |
| |
= |
f '(x) |
g(x) |
- |
f(x) |
g '(x) |
|
| ( g(x) )2 |
| |
= |
( ex + p )' |
( sin((p)*x) ) |
- |
( ex + p ) |
( sin((p)*x) )' |
|
| ( sin((p)*x) )2 |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( ex + p )' |
= |
ex + p (1 + 0) |
(by the chain rule) |
| ( sin((p)*x) )' |
= |
(p)*cos((p)*x) |
(by the chain rule) |
so the finished derivative is
| B' |
= |
( ex + p (1 + 0) ) |
( sin((p)*x) ) |
- |
( ex + p ) |
( (p)*cos((p)*x) ) |
|
| ( sin((p)*x) )2 |
| |
= |
ex + p sin((p)*x) - (p)*ex + p cos((p)*x) |
|
| (sin((p)*x))2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Sat Apr 4 21:46:48 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.