Thomas Calculus 13th [Solutions]

Section 3.5 Derivatives of Trigonometric Functions 147
36.
2
y tan x y sec x slope of tangent at x is
sec
2
3
4; slope of tangent at
2
3 3
2
x 0 is sec (0) 1; and
slope of tangent at x is sec 4. The tangent
at , tan , 3 is y 3
3 3 3
tangent at (0, 0) is y x ; and the tangent at
, tan , 3 is y 3 4 x .
3 3 3 3
3
4 x ; the
3
37. y sec x y sec x tan x slope of tangent at
x is sec tan 2 3; slope of tangent
3 3 3
at x issec tan 2. The tangent at the point
4 4 4
, sec , 2 is y 2 2 3 x ; the
3 3 3 3
tangent at the point
y
2 2 x .
4
, sec , 2 is
4 4 4
38. y 1 cos x y sin x slope of tangent at
3
3 2
sin ; slope of tangent at x
3
is
3
The tangent at the point
, 1 cos ,
3
3 3
3 3
2 2 3
3
,1 cos
3 3
, 1 is y 1 x
3
2 2 2 2
is y x ; the tangent at the point
2
x is
3
sin 1.
39. Yes, y x sin x y 1 cos x ; horizontal tangent occurs where 1 cos x 0 cos x 1 x
3 2
2
40. No, y 2x sin x y 2 cos x ; horizontal tangent occurs where 2 cos x 0 cos x 2. But there are
no x -values for which cos x 2.
41.
2
2 2
No, y x cot x y 1 csc x ; horizontal tangent occurs where 1 csc x 0 csc x 1. But there are no
2
x -values for which csc x 1.
42. Yes, y x 2 cos x y 1 2 sin x ; horizontal tangent occurs where 1 2 sin x 0 1 2sin x
1
sin x x or x
5
2 6 6
43. We want all points on the curve where the tangent
2
line has slope 2. Thus, y tan x y sec x so that
2
y 2 sec x 2 sec x 2 x Then the
tangent line at
tangent line at
4 .
4 , 1 has equation y 1 2 x ; the
4
4 , 1 has equation y 1 2 x .
4
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