MATH 100, HOMEWORK 3 SOLUTIONS
ex
Section 1.6, #26: If y =
then (1 + 2ex )y = ex , y + 2ex y = ex ,
1 + 2ex y y
, x = ln
. Hence the inverse function is
ex (1 − 2y) = y, ex =
1
−
2y
1
−
2y
x y = ln
.
1 − 2x
√
√
Section 1.6, #62: (a) Let x = cot−1 ( 3), then cot x = 3 and x ∈ (0, π),
hence x = π/6.
(b) Let x = arccos(− 21 ), then cos x = − 21 and x ∈ [0, π], hence x = 2π/3.
Section 3.5, #18: Differentiate the equation:
sec2 (x − y) · (1 − y 0 ) =
y 0 sec2 (x − y) +
0
y =
y 0 (1 + x2 ) − y · 2x
,
(1 + x2 )2
1 2xy
= sec2 (x − y) +
,
2
1+x
(1 + x2 )2
2xy
(1+x2 )2
1
+ 1+x
2
sec2 (x − y) +
sec2 (x − y)
.
Section 3.5, #42: We first find the equation of the tangent line at a point
(a, b). Differentiate the equation of the curve:
√
√
y
1 0
1/2 x
1
0
√ + √ y = 0, y = −
√ = −√ .
2 x 2 y
1/2 y
x
q
Thus at (a, b) the slope of the tangent is − b/a, and the equation of the
tangent is
s
b
y−b=−
(x − a).
a
We now find the intercepts.
s
x=0: y =b+
s
y=0:
√
b
· a = b + ab,
a
√
√
b
(x − a) = b, x − a = ab, x = a + ab.
a
1
The sum of the intercepts is
√
√
√
√
√
(b + ab) + (a + ab) = a + 2 ab + b = ( a + b)2 = c,
as required. At the last step we used the equation of the curve.
Section 3.5, #58: (a) If y = sec−1 x then x = sec y. Differentiate this:
1 = (sec y)0 = tan y sec y · y 0 , so that
y0 =
1
.
sec y tan y
To write the last expression in terms of x,√we use the identity tan2 y = sec2 y−
1 (derived in class). We have tan y = ± sec2 y − 1. If either 0 ≤ y < π/2 or
π ≤ y < 3π/2, we have tan y ≥ 0, hence we choose the + sign. Thus
y0 =
1
1
.
= √ 2
2
sec y sec y − 1
x x −1
√
(b) The first part of the derivation is the same, up to the point where we
have to choose √
the + or − sign. For 0 ≤ y < π/2, we have tan y ≥ 0,
hence tan√
y = sec2 y − 1. For π/2 < y ≤ π we have tan y ≤ 0, hence
tan y = − sec2 y − 1. Thus
0
y =





√1
,
sec2 y−1
√1
,
− sec y sec2 y−1
sec y
if 0 ≤ y < π/2;
if π/2 < y < π
But on the other hand,
| sec y| =
sec y,
if 0 ≤ y < π/2;
.
− sec y, if π/2 < y < π
Thus in both cases,
y0 =
1
1
√ 2
= √ 2
.
| sec y| sec y − 1
|x| x − 1
Section 3.6, #24:
y0 =
x−1 x2 − ln x · 2x
1 − 2 ln x
=
,
4
x
x3
2
.
y 00 =
−2x−1 x3 − (1 − 2 ln x) · 3x2
−5 + 6 ln x
=
.
x6
x4
Section 3.6, #46: We have ln y = ln x sin x, hence
1
1
y0
= ln(sin x) + ln x
cos x,
y
x
sin x
y0 = y
ln(sin x)
x
+ ln x cot x = (sin x)ln x
3
ln(sin x)
x
+ ln x cot x .