MATH1007*W
ANSWERS TO TUTORIAL 5
Winter 2004
µ
¶2
arcsin x
arcsin x
1. Simplifying cos
+ sin
, we obtain
2
2
(a) 1 +
p
(b)
1+x
1 + x + 2x2
(c)
1
1
+ x + x2
2
2
(d)
Answer. (b) 1 + x.
x(1 − x)
(e)
3
1
+ x + x2 .
2
2
For convenience, we let θ = (arcsin x)/2. Then sin 2θ = x and
µ
arcsin x
arcsin x
+ sin
cos
2
2
¶2
= (cos θ + sin θ)2
= cos2 θ + sin2 θ + 2 sin θ cos θ = 1 + sin 2θ = 1 + x.
2. The inverse function of u = arcsin
(a) w =
√
sin u
√
w is given by
(b) w = sin
√
(d) w = sin2 u
Answer. (d) w = sin2 u.
(c) w = sin(u2 )
u
(e) w =
Indeed, u = arcsin
√
p
sin(u2 ).
w gives sin u =
√
w and hence
w = (sin u)2 ≡ sin2 u.
3. The derivative of arctan (sin x) is
(a)
Answer. (e)
cos x
.
1 + sin2 x
1
1 + x2
(b)
(d) −
sin x
1 + sin2 x
Indeed,
x
1 + x2
(c)
(e)
sin x
1 − sin x
cos x
.
1 + sin2 x
1
cos x
d
arctan(sin x) =
cos x =
.
2
dx
1 + (sin x)
1 + sin2 x
1
4. The derivative of cos(arcsin
(a)
√
√
x) is
1
(b) − √
2 x
1
1−x
r
(d) −
x
1−x
1
(c) − √
2 1−x
1
(e) − p
x(1 − x2 )
.
1
. Indeed,
Answer. (c) − √
2 1−x
√
√
d
1
1
cos(arcsin x) = − sin(arcsin x) p
√ 2 √
dx
1 − ( x) 2 x
√
1
1
1
√ =− √
= − x√
.
1−x 2 x
2 1−x
5. The derivative of y = ln(sec x + tan x) is
(a)
sec x
(b)
1
sec x + tan x
(d) tan x
Remark:
(e)
(c)
sec x − tan x
sec x + tan x
tan x
.
sec x + tan x
d
1
d x
ln x =
and
e = ex .
dx
x
dx
Answer. (a) sec x. Indeed,
d
1
ln(sec x + tan x) =
(sec x tan x + sec2 x)
dx
sec x + tan x
1
=
sec x(sec x + tan x) = sec x.
sec x + tan x
2
6. Find the slope of the tangent line to the curve described implicitly as
e2x − esin x + ln y = 0
at the point (0, 1) on the curve.
(a) y = 1
(b) x = 0
Answer. (d) y = −x + 1.
(c) y = ex + 1
(d) y = −x + 1
(e) y = −2x + 1.
Differentiating both sides of e2x − esin x + ln y = 0, we get
2e2x − esin x cos x +
1 0
y = 0.
y
Letting x = 0 and y = 1, we get 2 − 1 + y 0 = 0 and hence y 0 = −1. This tells us
that the required tangent has slope −1. Thus the tangent line is y = −x + 1.
3