Review of Derivatives, Integrals, Tangent Lines
Solutions Manual
Math 126E. 03/29/2011
(a) Recall: (sec x)0 = tan x sec x.
Problem 1.
3 sec3 x tan x,
f 0 (x) =
Therefore, (sec3 x)0 = 3 sec2 x · tan x sec x =
(sec3 x)0 (1 + x2 ) − sec3 x(1 + x2 )0
3 sec3 x tan x(1 + x2 ) − 2x sec3 x
=
.
(1 + x2 )2
(1 + x2 )2
f 0 (x) is defined iff sec x = 1/ cos x and tan x = sin x/ cos x are defined, i.e. iff cos x 6= 0, x 6= π/2 + πn,
where n ∈ Z (n is an integer).
(b)
dx3
d
dy
2
2
2
2
2
2
2
2
=
cos(ex ) + x3 cos(ex ) = 3x2 cos(ex ) + x2 · 2xex (− sin(ex )) = 3x2 cos(ex ) − 2x4 ex sin(ex ).
dx
dx
dx
(c)
(a tan(bv))0v
ab sec2 (bv)
b cos−2 (bv)
b
du
=
=
=
=
.
dv
a tan(bv)
a tan(bv)
sin(bv)/ cos(bv)
sin(bv) cos(bv)
(d)
dy
1
d2 y
d
= 2
,
=
dt
t + 1 dt2
dt
1
t2 + 1
=−
2t
.
(1 + t2 )2
Problem 2. (a) Let t = ln x. Then dt = dx/x, and the segment [e, e2 ] of integration changes to
[1, 2]. So
Z2
Ze2
dt
dx
=
.
I=
x(ln x)p
tp
1
e
If p = 0, then I =
R2
1
dt = 1. If p = 1, then I =
Z2
I=
1
R2
1
dt/t = log t|t=2
t=1 = log 2. If p 6= 1, then
t=2
t1−p 21−p − 1
t dt =
=
.
1 − p t=1
1−p
−p
(b) Integration by parts: u = x, v = −e−2x /2, v 0 = e−2x .
Z
Z
Z
Z −2x e−2x
e
−2x
0
0
xe dx = uv dx = uv − u vdx = −x
−
dx =
2
2
Z
e−2x 1
xe−2x 1 −2x
−x
+
e−2x dx = −
− e
+ C.
2
2
2
4
(c)
Z∞
2/3
dx
1
=
2
9x + 4
9
Z∞
dx
1
=
2
x + 4/9
9
Z∞
1
2/3
1
2
dt
3
2 2
( 3 t) +
1
.
Here, we have made the change of variables: t = 3/2x, x = 2/3t, dx = 2/3dt. This expression equals
12
93
Z∞
1
dt
129
=
4 2
934
(t + 1)
9
Z∞
1
dt
1
=
2
t +1
6
Z∞
t2
dt
=
+1
1
1
1 π π 1
· arctan t|t=∞
=
−
=
.
t=1
6
6 2
4
24π
Problem 3. f 0 (x) = − 12 (1 − x)−1/2 . f 0 (0) = −1/2, f (0) = 1. Tangent line: y − f (0) = f 0 (0)(x − 0),
√
y = 1 − 1/2x. Therefore, 0.99 = f (0.01) ≈ 1 − 0.01/2 = 0.995.
2