1 In Class Questions MATH 151-Fall 02 October 9 1. Use

1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
October 9
1. Use the table of information to compute the following. If it is not possible explain why.
x
f (x)
f 0 (x)
2
3
6
3
4
7
4
5
8
5
2
9
x
g(x)
g0 (x)
J(x) = f (g(x)),
2
1
−5
3
2
−6
4
5
−7
5
8
−9
H(x) = g(x + f (x))
(a) J 0 (x) = f 0 (g(x)) ∗ g0 (x) so J 0 (4) = f 0 (g(4)) ∗ g0 (4) = f 0 (5) ∗ −7 = 9 ∗ −7 = −63
(b) H 0 (x) = g0 (x+f (x))∗(1+f 0 (x)) so H 0 (2) = g0 (2+f (2))∗(1+f 0 (2)) = g0 (2+3)∗(1+6) =
−9 ∗ 7 = −63
2. Compute these trig limits.
(a) lim
x→0
lim
x→0
1 − cos(5x)
1 − cos(5x) 1 + cos(5x)
1 − cos2 (5x)
sin2 (5x)
=
lim
∗
=
lim
=
lim
=
x→0
6x2
6x2
1 + cos(5x) x→0 6x2 (1 + cos(5x)) x→0 6x2 (1 + cos(5x))
5 ∗ sin(5x) 5 ∗ sin(5x)
1
1
25
∗
∗
=5∗5∗
=
5x
5x
6(1 + cos(5x))
12
12
(b) lim
sin(5x)
sin(5x) cos(3x)
5 ∗ sin(5x)
3x
1
5
= lim
= lim
∗
∗ cos(3x) = 5 ∗ ∗ 1 =
x→0
tan(3x) x→0
sin(3x)
5x
3 ∗ sin(3x)
3
3
(c) lim
tan(4x)
= lim
3 tan(7x) x→0
x→0
x→0
sin(4x)
cos(4x)
3 sin(7x)
cos(7x)
= lim
x→0
sin(4x) cos(7x)
=
3 sin(7x) cos(4x)
4 ∗ sin(4x)
7x
cos(7x)
1 1
4
∗
∗
=4∗ ∗ =
x→0
4x
7 ∗ sin(7x) 3 cos(4x)
7 3
21
tan(x)
1
(d) limπ
=
x→ 4
4x
π
lim
(e) lim
x→1
sin(x − 1)
sin(x − 1)
sin(x − 1)
1
1
1
= lim
= lim
∗
=1∗ =
x2 + x − 2 x→1 (x − 1)(x + 2) x→1 (x − 1)
(x + 2)
3
3
3. Take the derivatives of these functions.
cos(x)
(a) y = sin
x2 + 1
"
cos(x)
−(x2 + 1) sin(x) − cos(x) ∗ 2x
y = cos
∗
x2 + 1
(x2 + 1)2
0
#
(b) y = sec2 (x4 + 1) = (sec(x4 + 1))2
y 0 = 2(sec(x4 + 1)) ∗ sec(x4 + 1) tan(x4 + 1) ∗ 4x3
(c) y = tan(5x) cos(x2 + 5)
y 0 = 5 sec2 (5x) ∗ cos(x2 + 5) + tan(5x) ∗ (− sin(x2 + 5) ∗ 2x)
(d) y = sin2 (cos(4x)) = (sin(cos(4x)))2
y 0 = 2(sin(cos(4x))) ∗ cos(cos(4x)) ∗ (− sin(4x)) ∗ 4