Derivatives of Constant Multiples
Example:
f(t) = (((0)4))*sqrt(t4 - t3 + 2*t)
The first thing to notice when finding the derivative of this function
is that it is
the product of a constant and another function,
as shown in color below:
| f(t) |
= |
((0)4) |
sqrt(t4 - t3 + 2*t) |
The Derivative Rule for Constant Multiples:
The derivative of a constant multiple is the constant times thederivative of the function.
If
then the derivative of
z is
| |
z' |
= |
( c |
f(x) )' |
| |
|
= |
c |
f '(x) |
So our example,
| f(t) |
= |
((0)4) |
sqrt(t4 - t3 + 2*t) |
we can think of as
So the derivative is
| f '(t) |
= ( |
c |
g(t) |
)' |
| |
= |
c |
g '(t) |
|
| |
= |
((0)4) |
(sqrt(t4 - t3 + 2*t))' |
|
and we just need to know the derivative on the right-hand
side of the equation. In this case this is
| sqrt(t4 - t3 + 2*t) |
= |
(1/2)*(t4 - t3 + 2*t)-1/2 (4*t3 - 3*t2 + 2) |
(by the chain rule) |
so the finished derivative is
| f '(t) |
= |
((0)4) |
( (1/2)*(t4 - t3 + 2*t)-1/2 (4*t3 - 3*t2 + 2) ) |
| |
= |
(1/2(((0)4)))*(t4 - t3 + 2*t)-1/2 (4*t3 - 3*t2 + 2) |
![[]](/images/boxby.gif)
additional explanation for the derivative of constant multiples
see another derivative of constant multiples example
practice gateway test
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Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.