Section 3.5 Homework Exercises
1.
d
dx
(−10x + 3 cos x) = −10 − 3 sin x
3.
d
dx
x2 cos x = x2
7.
d
dx (sin
d
dx
cos x +
x tan x) = sin x
d
dx
d
dx
x2 = −x2 sin x + 2x cos x
tanx + tan x
d
dx
sin x = sin x sec2 x + tan x cos x
13. A problem like this will be a candidate for other rules later on once we have them, but for
now we employ the quotient rule.
d
dx
4
1
+
cos x tan x
19. For s = tan t − t, we have
ds
dt
d
dx
d
d
4 − 4 dx
cos x tan x dx
1 − tan x
+
2
cos x
tan2 x
(cos x)(0) − 4(−sin x) (tan x)(0) − 4 sec2 x
=
+
cos2 x
tan2 x
4 sin x 4 sec2 x
−
=
cos2 x
tan2 x
sin x
1
1
cos2 x
=4·
·
−4·
·
cos x cos x
cos2 x sin2 x
= 4 tan x sec x − 4 csc2 x
=
cos x
= sec2 t − 1.
23. If r = 4 − θ2 sin θ, then
dr
d
d
= −θ2
sin θ + sin θ
(−θ2 )
dθ
dθ
dθ
= −θ2 cos θ − 2θ sin θ
29. For the derivative of p =
sin q+cos q
,
cos q
we need to use the quotient rule.
d
d
(sin q + cos q) − (sin q + cos q) dq
cos q
cos q dq
dp
=
2
dq
cos q
cos q(cos q − sin q) − (sin q + cos q)(−sin q)
=
cos2 q
cos2 q − sin q cos q + sin2 q + sin q cos q
=
cos2 q
cos2 q + sin2 q
=
cos2 q
1
=
cos2 q
= sec2 q
35. For y = sin x,
dy
dx
= cos x.
y =0
x=−π
dy = −1
dx x=−π
0 = −(−π) + b
y =0
x=0
dy =1
dx x=0
0 = 1(0) + b
b = −π
b=0
y dy =0
dx x=−π
3π
+b
−1 = 0
2
b = −1
y=x
y = −x − π
= −1
x= 3π
2
y = −1
y=x
y = −x − π
y = −1
39. To find if there are any horizontal tangents to the curve, we find y 0 and set equal to 0.
y = x + sin x
0
y = 1 + cos x = 0
cos x = −1
x=π
47. lim sin
x→2
1
x
−
1
2
= sin
59. We want to find
1
2
−
d999
cos
dx999
1
2
= sin 0 = 0
x.
d
cos x = −sin x
dx
d2
d
cos x =
(−sin x) = −cos x
dx2
dx
d3
d
− cos x = sin x
cos x =
dx3
dx
d4
d
sin x = cos x
cos x =
dx4
dx
Notice that it takes 4 derivatives to return back to cos x. When we find 999
4 , we 249 with a
remainder of 3. This tells us that we would go through this cycle 249 times and then would
be back at cos x needing to take 3 more derivatives. So,
d999
d3
cos
x
=
cos x = sin x
dx999
dx3