17. ((2x−3)4 (x2 +x+1)5 )0 = 4(2x−3)3 ·2·(x2 +x+1)5 +(2x−3)4 ·5(x2 +x+1)4 ·(2x+1) =
(2x − 3)3 (x2 + x + 1)4 (28x2 − 12x − 7).
31. (sin(tan 2x))0 = cos(tan 2x) · sec2 (2x) · 2 = 2 cos(tan 2x) sec2 (2x).
2
2
2
2
2
2
2
41. (sin2 (esin t ))0 = 2 sin(esin t ) cos(esin t )·esin t ·2 sin t cos t = 4 sin(esin t ) cos(esin t )esin t sin t cos t.
63. (a) (f ◦ g)0 (1) = f 0 (g(1)) · g 0 (1) = 5 · 6 = 30.
(b) (g ◦ f )0 (1) = g 0 (f (1)) · f 0 (1) = 9 · 4 = 36.
65. (a) (f ◦ g)0 (1) = f 0 (g(1)) · g 0 (1) = f 0 (3) · (−3) = 43 .
(b) (g ◦ f )0 (1) = g 0 (f (1)) · f 0 (1) = g 0 (2) · 2 DNE since g 0 (2) DNE.
(c) (g ◦ g)0 (1) = g 0 (g(1)) · g 0 (1) = g 0 (3) · (−3) = −2.
69. (a) F 0 (x) = f 0 (ex ) · (ex )0 = ex f 0 (ex ).
(b) G0 (x) = ef (x) f 0 (x).
75. y 0 = e2x (−3A sin 3x + 3B cos 3x) + 2e2x (A cos 3x + B sin 3x) = e2x ((2A + 3B) cos 3x +
(2B − 3A) sin 3x). y 00 = e2x ((2(2A + 3B) + 3(2B − 3A)) cos 3x + (2(2B − 3A) − 3(2A +
3B)) sin 3x) = e2x ((12B − 5A) cos 3x + (−12A − 5B) sin 3x).
78. (xe−x )0 = e−x − xe−x = (1 − x)e−x . ((1 − x)e−x )0 = −e−x − (1 − x)e−x = −(2 − x)e−x .
(−(2 − x)e−x )0 = e−x + (2 − x)e−x = (3 − x)e−x . Claim: (xe−x )(n) = (−1)n−1 (n −
x)e−x . By induction on n, only need to consider ((−1)n−1 (n − x)e−x )0 = (−1)n e−x +
(−1)n−1 (n − x)(−e−x ) = (−1)n (n + 1 − x)e−x . Hence, n = 1000 ⇒ −(1000 − x)e−x .
2π
= 0.35 cos 2πt
· 5.4 = 7π
cos 2πt
.
81. (a) dB
dt
5.4
54
5.4
(b) t = 1 ⇒≈ 0.16.
86. (a) Rate of change of the volume of the balloon w.r.t. its radius or the time, respectively.
2
= 34 π d(rdt ) = 83 π dr
.
(b) V = 43 πr2 ⇒ dV
dt
dt
x
x
dy
y·1−x· dx
dy
de y
15. dx = e y
and d(x−y)
= 1 − dx
. Hence, we have
2
y
dx
x
ey
−1
y
x
y
xe
−1
y2
x
ey
y
x
y
− xey2
dy
dx
dy
= 1 − dx
⇒
dy
dx
=
x
=
ye y −y 2
x
xe y −y 2
.
23. (4x3 dx
y 2 + x4 (2y)) − (3x2 dx
y + x3 ) + (2 dx
y 3 + 2x(3y 2 )) = 0 ⇒ (4x3 y 2 − 3x2 y + 2y 3 ) dx
=
dy
dy
dy
dy
(−2x4 y + x3 − 6xy 2 ) ⇒
dx
dy
=
−2x4 y+x3 −6xy 2
.
4x3 y 2 −3x2 y+2y 3
31. 4(x2 + y 2 ) · (2x + 2y 0 ) = 25(2x − 2yy 0 ). For x = 3, y = 1, we have 4(9 + 1) · (6 + 2y 0 ) =
25(6 − 2y 0 ) ⇒ y 0 = −9
. Tangent line: y − 1 = −9
(x − 3).
13
13
1
43. Horizontal ⇔ y 0 = 0 ⇒ 4(x2 + y 2 ) · (2x) = 25(2x). Then, either x = 0 or x2 + y 2 = 25
.
4
However, the only point on lemniscate with x = 0 is (0, 0), which gives no information
about y 0 via the equation 4(x2 + y 2 ) · (2x + 2y 0 ) = 25(2x − 2yy 0 ). √
And x2 + y 2 = 25
4
2
5 3
25
5
2
2
=
25(x
−
+
x
)
⇒
x
=
±
and
y
=
±
.
gives you the correct points: 2 · 25
4
4
4
4
√
dy
dy
dy
−1
63. Suppose y = cos−1 x ⇔ x = cos y ⇒ 1 = − sin y dx
= − 1 − x2 dx
⇒ dx
= √1−x
2.
√
2
Note. Again, why sin y should be positive rather than being − 1 − x ?
2