Derivatives of Products
Example:
f(y) = sqrt(y4 + 1) ln((1/2)*y)
The first thing to notice when finding the derivative of this function
is that it is
the product of several terms,
as shown in color below:
| f(y) |
= |
( sqrt(y4 + 1)) |
( ln((1/2)*y)) |
The Derivative Rule for Products:
The derivative of a product is the derivative of the first term times the second (and third, etc.) term(s), plus the first (and third, etc.) term(s) times the derivative of the second, etc.
If
then the derivative of
z is
| |
z ' |
= |
(f(x) g(x))' |
| |
|
= |
f '(x) g(x) |
+ |
f(x) g '(x) |
So our example,
| f(y) |
= |
( sqrt(y4 + 1)) |
( ln((1/2)*y)) |
we can think of as
So the derivative is
| f '(y) |
= ( |
g(y) |
h(y) |
)' |
| |
| |
= |
g '(y) |
h(y) |
+ |
g(y) |
h '(y) |
| |
= |
( sqrt(y4 + 1) )' |
( ln((1/2)*y) ) |
+ |
( sqrt(y4 + 1) ) |
( ln((1/2)*y) )' |
and we just need to know each of the derivatives on the right-hand
side of the equation. In this case these are
| ( sqrt(y4 + 1) )' |
= |
(1/2)*(y4 + 1)-1/2 (4*y3 + 0) |
(by the chain rule) |
| ( ln((1/2)*y) )' |
= |
(1/2)*((1/2)*y)-1 |
(by the chain rule) |
so the finished derivative is
| f '(y) |
= |
( (1/2)*(y4 + 1)-1/2 (4*y3 + 0) ) |
( ln((1/2)*y) ) |
+ |
( sqrt(y4 + 1) ) |
( (1/2)*((1/2)*y)-1 ) |
| |
= |
2*y3 (y4 + 1)-1/2 ln((1/2)*y) + (1/2)*sqrt(y4 + 1) ((1/2)*y)-1 |
![[]](/images/boxby.gif)
additional explanation for the product rule
see another product rule example
practice gateway test
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Comments to Gavin LaRose
glarose@umich.edu
©2001 Gavin LaRose,
University of Michigan Math Dept.