Lecture Note Differentiation
Example 1
Suppose that yu u=+ andux=+^317. Use the Chain Rule to find
dy
dx
, then
evaluate
dy
dx
a n is denoted by
x 2
dy
dx
t x=2 (Such an evaluatio
=
)
Solution
yu u uu=+ =+½ and ux=+^317
1
1112
2
+u 1 a
2
dy
du u
−
= =+ nd^32
du
x
dx
=
So, by the Chain Rule,
131 2
2
dy dy du
x
dx du
==
dx u
⎛⎞
⎜⎟+
⎝⎠
and
Ifx= 2 , thenu=+ 217253 = 32 12^2
du
dx
= ×=. Hence
2
1 6
12
⎞
⎟×=
16
1121
x^225105
dy
dx =
⎛⎞⎛
=+ ×=+⎜⎟⎜
⎝⎠⎝⎠
Example 2
Find
dy
when 1
dx
x= if
1
u
y= and ux 32
u+
= − 1 (Answer: 2/3)
en e function
Example 3
Differ tiate th
()( )
4 3
a.fx()=+x^232 x+ b.f xxx=− 2 c. ()
()
5
1
23
fx
x
=
+
Example 4
An environmental study of a certain suburban community suggests that the average
daily level of carbon monoxide in the air will be Cp()= 0.5p^2 + 17 parts per million
when the population is p thousand. It is estimated that t years from now, the
population of the community will be p() 3.1t=+0.1^2
At what rate will the carbon monoxide level be changing with respect to tim
from now?
t thousand.
e 3 years
is to find
Solution
dC
dt
The goal whent= 3. Since
(^) ()() ()
11
(^11221722) 2. 0.5 17
22
ppp
dp
0.5 0.5
dC
p
− −
=+⎡⎣⎤⎦= +
and
0.2
dp
t
dt
=
it follows from the chain rule that
()()
1
(^22)
2
10 .1
0.5 17 0.2
0.5 17
dc dc dp pt
p t
p
−
+ =
2
p
dt dp dt