Derivatives of Quotients
Example:
y = sin(x + 1) / ln(x)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| y |
= |
sin(x + 1) |
 |
| ln(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,
| y |
= |
sin(x + 1) |
|
| ln(x) |
we can think of as
| y |
= |
f(x) |
|
| g(x) |
So the derivative is
| y' |
= ( |
f(x) |
)' |
|
| g(x) |
| |
= |
f '(x) |
g(x) |
- |
f(x) |
g '(x) |
|
| ( g(x) )2 |
| |
= |
( sin(x + 1) )' |
( ln(x) ) |
- |
( sin(x + 1) ) |
( ln(x) )' |
|
| ( ln(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
so the finished derivative is
| y' |
= |
( cos(x + 1) (1 + 0) ) |
( ln(x) ) |
- |
( sin(x + 1) ) |
( x-1 ) |
|
| ( ln(x) )2 |
| |
= |
cos(x + 1) ln(x) - x-1 sin(x + 1) |
|
| (ln(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 Jul 5 21:20:38 2025
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.