September 15, 2014
1
(a)
E(x) = 12 (f (x) + f (−x)) =
O(x) = 21 (f (x) − f (−x)) =
x4
1−x2
x5
− 1−x
2
(b)
See figures in Appendix.
2
g 00 (x) = (f 2 (x))00 = (2f (x)f 0 (x))0 = 2(f 0 (x))2 + 2f (x)f 00 (x)
Thus g 00 (x0 ) = 2(f 0 (x0 ))2 + 2f (x0 )f 00 (x0 ) > 0
3
(a)
d
2
−3
dx (arctan(x + x + 1))
d
2
arctan(x2 + x + 1)
= −3(arctan(x + x + 1))−4 dx
1
d
2
= −3(arctan(x2 + x + 1))−4 1+(x2 +x+1)
2 dx (x + x + 1)
1
= −3(arctan(x2 + x + 1))−4 1+(x2 +x+1)
2 (2x + 1)
(b)
d
dy sin y cos y
= (sin y)0 cos y + sin y(cos y)0
= cos2 y − sin2 y = cos 2y
d
dy
sin y cos y
d 1
= dy
2 sin 2y
1
= 2 cos 2y · 2
= cos 2y
1
4
(a)
d( 13 y 3 + xy 2 + x2 y + x3 + x − 53 )
= y 2 dy + y 2 dx + 2xydy + 2xydx + x2 dy + 3x2 dx + dx
= (y 2 + 2xy + 3x2 + 1)dx + (y 2 + 2xy + x2 )dy
Set y 2 + 2xy + 3x2 + 1 = 0. Thus we see that there isn’t any point at which
the tangent line is horizontal.
dy
dx
= 0, i.e.
(b)
Set x2 + 2xy + y 2 = 0. We see that y=-x. Plugging this into the equation, we have
5
2 3
3 x + x = 3 . x = 1 is one solution to it. Thus at the point (1, −1) the tangent line is
vertical.
5
(a)
y = cos−1 (x)
cos y = x
dy
− sin y dx
=1
dy
1
√ 1
dx = − sin y = − 1−x2
(b)
sin(π/2 − cos−1 (x)) = cos(cos−1 (x)) = x
Thus π/2 − cos−1 (x) = sin−1 (x)
sin−1 (x) + cos−1 (x) = π/2
(c)
We know that sin−1 (x) + cos−1 (x) is a constant. Taking derivatives of both sides, we
d
d
see that dx
sin−1 (x) = − dx
cos−1 (x)
6
(a)
ln(uv) = ln u + ln v
(ln(uv))0 = (ln u)0 + (ln v)0
1
1 0
1 0
0
uv (uv) = u u + v v
This implies (uv)0 = uv 0 + u0 v
(b)
(ln( uv ))0 = (ln u)0 − (ln v)0
v u 0
1 0
1 0
u(v ) = uu − vv
0
( uv )0 = v1 u0 − uv
v2
2
( uv )0 =
u0 v−uv 0
v2
(c)
ln(u1 u2 ...un ) = ln u1 + ln u(2) + ... + ln u(n)
(ln(u1 u2 ...un ))0 = (ln u1 )0 + (ln u2 )0 + ... + (ln un )0
1
1 0
1 0
1 0
0
u1 u2 ...un (u1 u2 ...un ) = u1 u1 + u2 u2 + ... + un un
(u1 u2 ...un )0 = u01 u2 ...un + u1 u02 ...un + ... + u1 u2 ...u0n
7
(8a)
1
M1 = 32 log10 E
E0 , M2 =
1 = M2 − M1 = 23 log10
E2
3/2 ≈ 31.62
E1 = 10
E2
2
3 log10 E0
E2
E1
(8c)
Energy released by an earthquake of magnitude 6 is (103/2 )6 E0 = 109 E0 = 7 × 106
The number of days is (7 × 106 )/(3 × 105 ) ≈ 23.3
(10)
p
Assume log3 2 = pq , p and q are positive integers. Then we have 3 q = 2. Thus 3p = 2q .
This is impossible because 3p is odd while 2q is even.
(11)
The first inequality is true only when the base is between 0 and 1, while the last step is
true only when the base is bigger than 1. Thus no matter what the base is, one of the
two steps is false.
(18)
ln y = 13 (ln(x + 1) + ln(x − 2) + ln(2x + 7)) Differentiating this equation with respect to
dy
dy
1
1
2
1
1
2
x, we get y1 dx
= 13 ( x+1
+ x−2
+ 2x+7
). Thus we have dx
= y3 ( x+1
+ x−2
+ 2x+7
)
(20)
y = ex ln x
y 0 = ex ln x (ln x + 1)
ln y = x ln x
1 0
y y = ln x + 1
y 0 = y(ln x + 1)
8 Appendix
3
x4/(1−x2)
10
8
6
4
2
0
−2
−4
−6
−8
−10
−2
−1.5
−1
−0.5
0
x
0.5
1
1.5
2
1
1.5
2
Figure 1: E(x)
4 x3/(1−x2)2
10
8
6
4
2
0
−2
−4
−6
−8
−10
−2
−1.5
−1
−0.5
0
x
0.5
Figure 2: E 0 (x)
4
−x5/(1−x2)
10
8
6
4
2
0
−2
−4
−6
−8
−10
−2
−1.5
−1
−0.5
0
x
0.5
1
1.5
2
1
1.5
2
Figure 3: O(x)
x4 (3 x2−5)/(1−x2)2
10
8
6
4
2
0
−2
−4
−6
−8
−10
−2
−1.5
−1
−0.5
0
x
0.5
Figure 4: O0 (x)
5