Derivatives of Quotients
Example:
W = e6*z / sqrt(z)
The first thing to notice when finding the derivative of this function
is that it is a quotient, as shown below:
| W |
= |
e6*z |
 |
| sqrt(z) |
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,
| W |
= |
e6*z |
|
| sqrt(z) |
we can think of as
| W |
= |
f(z) |
|
| g(z) |
So the derivative is
| W' |
= ( |
f(z) |
)' |
|
| g(z) |
| |
= |
f '(z) |
g(z) |
- |
f(z) |
g '(z) |
|
| ( g(z) )2 |
| |
= |
( e6*z )' |
( sqrt(z) ) |
- |
( e6*z ) |
( sqrt(z) )' |
|
| ( sqrt(z) )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
| W' |
= |
( 6*e6*z ) |
( sqrt(z) ) |
- |
( e6*z ) |
( (1/2)*z-1/2 ) |
|
| ( sqrt(z) )2 |
| |
= |
6*e6*z sqrt(z) - (1/2)*z-1/2 e6*z |
|
| (sqrt(z))2 |
![[]](/images/boxby.gif)
additional explanation for the quotient rule
see another quotient rule example
practice gateway test
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Page Generated: Fri Jan 23 20:27:06 2026
Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.