3.5 SOLUTIONS
2.
4.
6.
8.
10.
12.
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
= −3x−2 + 5 cos x
= 12 x−1/2 sec x + x1/2 sec x tan x
= 2x cot x − x2 csc2 x + 2x−3
= − csc x cot2 x − csc3 x
= (cos x − sin x) sec x + (sin x cos x) sec x tan x
dy
(1 + sin x)(− sin x) − cos2 x
=
dx
(1 + sin x)2
− sin x − sin2 x − cos2 x
=
(1 + sin x)2
−1 − sin x
=
(1 + sin x)2
−1
=
1 + sin x
dy
= − tan2 x + (2 − x)(2 tan x sec2 x)
dx
dr
24.
= sin θ + θ cos θ − sin θ
dθ
dp
(1 + tan q) sec2 q − tan q sec2 q
30.
=
dq
(1 + tan q)2
dy
38. We are given y = 1 + cos x, so
= − sin x. To find the tangent line, we need the
dx
dy
slope and a point. To get the slope, we plug in the x-value in to
.
dx
√
• x = − π3 : The slope is − sin(− π3 ) = 23 . √The point is (− π3 , 1+cos(− π3 )) = (− π3 , 32 ).
Therefore the tangent line is y − 32 = 23 (x + π3 ).
• x = 3π
: The slope is − sin( 3π
) = 1. The point is ( 3π
, 1 + cos( 3π
) = ( 3π
, 1).
2
2
2
2
2
3π
Therefore the tangent line is y − 1 = x − 2 .
18.
1
2
3.5 SOLUTIONS
2 y
y = 1 + cos x
√
y − 23 = 23 (x + π3 )
y − 1 = x − 3π
2
1.5
1
0.5
x
−4
−2
2
4
6
48. We can use direct substitution in computing limits provided there is no division by
zero. Therefore
p
p
lim
1 + cos(π csc x) = 1 + cos(π csc(−π/6))
x→−π/6
p
= 1 + cos(−2π)
√
= 1+1
√
= 2.
50. Direct substitution would yield zero in the numerator and denominator. Thus we
need to simplify the expression algebraically first. Let x = θ − π4 . Then as θ → π4 ,
we have x → 0. We compute
tan(x + π4 ) − 1
tan θ − 1
=
lim
π
x→0
θ→π/4 θ −
x
4
tan(x + π4 ) − 1
1
= lim
−1
x→0
1−x
x
1
2 tan x
= lim
x→0
1 − tan x x
2 sin x
1
= lim
x→0
cos x − sin x x
sin x
2
= lim
x→0
x
cos x − sin x
2
=1·
1−0
= 2.
lim
trig identity
mult. by cos x
54. We compute
lim cos
θ→0
πθ
sin θ
πθ
= cos lim
θ→0 sin θ
= cos(π)
= −1.
3.5 SOLUTIONS
3
58. As we approach 0 from the left the slope of the tangent line is the slope of x + b,
which is 1, regardless of the value of b. From the right, the slope of the tangent line
is − sin(0) = 0. It follows that for any value of b, we will have a corner. In particular,
there is no value of b which makes g differentiable at x = 0.