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Math 124 Homework 18 (section 3.4), solutions to graded problems
20)
By two applications of the Chain Rule, we have
g
0
(
t
) =
e
(1+3
t
)
2
2(1 + 3
t
)3 =
6(1 + 3
t
)
e
(1+3
t
)
2
29)
w
0
=
3
2
(
x
2
·
5
x
)
1
2
(
x
2
(ln 5)5
x
+ 2
x
5
x
)
51)
f
0
(
x
) =
a
(
x
(

b
)
e

bx
+
e

bx
) =

abxe

bx
+
ae

bx
54)
This involves a “symbolic” use of the Chain Rule:
u
0
(
x
) =
g
0
(
f
(
x
))
f
0
(
x
)
(a)
u
0
(1) =
g
0
(
f
(1))
f
0
(1) =
g
0
(2)(2) =

1(2) =

2
(b)
u
0
(2) =
g
0
(
f
(2))
f
0
(2) =
undeﬁned
, since
f
0
(2) is undeﬁned.
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Unformatted text preview: 61) We ﬁrst diﬀerentiate: f ( x ) = 3( x1) 2 . Hence the slope of the tangent line at x = 2 is m = f (2) = 3(1) 2 = 3. Noting that f (2) = (21) 3 = 1 and using the pointslope form for a line, we have y1 = 3( x2) or y = 3 x5 1...
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This note was uploaded on 06/16/2008 for the course MATH 124 taught by Professor Kennedy during the Spring '08 term at University of Arizona Tucson.
 Spring '08
 KENNEDY
 Calculus, Chain Rule, The Chain Rule

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