Derivatives of Quotients
Example:
y = sqrt(t) / (t - p)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| y |
= |
sqrt(t) |
 |
| t - p |
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 |
= |
sqrt(t) |
|
| t - p |
we can think of as
| y |
= |
f(t) |
|
| g(t) |
So the derivative is
| y' |
= ( |
f(t) |
)' |
|
| g(t) |
| |
= |
f '(t) |
g(t) |
- |
f(t) |
g '(t) |
|
| ( g(t) )2 |
| |
= |
( sqrt(t) )' |
( t - p ) |
- |
( sqrt(t) ) |
( t - p )' |
|
| ( t - p )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' |
= |
( (1/2)*t-1/2 ) |
( t - p ) |
- |
( sqrt(t) ) |
( 1 - 0 ) |
|
| ( t - p )2 |
| |
= |
(1/2)*t-1/2 (t - p) - sqrt(t) |
|
| (t - p)2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Mon Dec 1 01:59:18 2025
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.