Derivatives of Sums
Example:
B(q) = sqrt(1 - q) + (sqrt(3))*q5 - 5
The first thing to notice when finding the derivative of this function
is that it is
the sum of several terms,
as shown in color below:
| B(q) |
= |
( sqrt(1 - q) ) | + |
( (sqrt(3))*q5 ) |
- |
( 5 ) |
The Derivative Rule for Sums:
The derivative of a sum is the sum of the derivatives.
If
then the derivative of
z is
| |
z' |
= |
( f(x) |
+ |
g(x) )' |
| |
|
= |
f '(x) |
+ |
g'(x) |
So our example,
| B(q) |
= |
( sqrt(1 - q) ) | + |
( (sqrt(3))*q5 ) |
- |
( 5 ) |
we can think of as
| B(q) |
= |
f(q) |
+ |
g(q) |
- |
h(q) |
So the derivative is
| B '(q) |
= ( |
f(q) |
+ |
g(q) |
- |
h(q) |
)' |
| |
= |
f '(q) |
+ |
g '(q) |
- |
h '(q) |
|
| |
= |
( sqrt(1 - q)) ' |
+ |
( (sqrt(3))*q5) ' |
- |
( 5) ' |
|
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
| B '(q) |
= |
(1/2)*(1 - q)-1/2 (0 - 1) |
+ |
(sqrt(3))*5*q4 |
- |
0 |
|
| |
= |
(-1/2)*(1 - q)-1/2 + (5(sqrt(3)))*q4 |
![[]](/images/boxby.gif)
additional explanation for the derivative of sums
see another derivative of sums example
practice gateway test
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Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.