Derivatives of Sums
Example:
y = t1/4 - (t3 - 1)-1
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:
| y |
= |
( t1/4 ) | - |
( (t3 - 1)-1 ) |
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,
| y |
= |
( t1/4 ) | - |
( (t3 - 1)-1 ) |
we can think of as
So the derivative is
| y ' |
= ( |
f(t) |
- |
g(t) |
)' |
| |
= |
f '(t) |
- |
g '(t) |
|
| |
= |
( t1/4) ' |
- |
( (t3 - 1)-1) ' |
|
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( t1/4 )' |
= |
(1/4)*t-3/4 |
(by the power rule) |
| ( (t3 - 1)-1 )' |
= |
(-1)*(t3 - 1)-2 (3*t2 - 0) |
(by the chain rule) |
so the finished derivative is
| y ' |
= |
(1/4)*t-3/4 |
- |
(-1)*(t3 - 1)-2 (3*t2 - 0) |
|
| |
= |
(1/4)*t-3/4 + 3*t2 (t3 - 1)-2 |
![[]](/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.