Math 155a_Exam Solution on Equations, Implicit Differentiation, Linear Approximation, Absolute Maxim

MATH 155A FALL 13
SOLUTIONS TO THE PRACTICE MIDTERM 2.
Question 1. Find y 0 .
(a) y =
1
√
x
-
1
5
√
x
3
.
(b) y =
tan x
1 + cos x
.
(c) y = x
√
21+ π
cos x.
(d) y =
1
sin( x - sin x )
.
(e) y = sin
2
(cos(
p
sin( πx )) .
Solution. Direct application of the differentiation rules yields:
(a) y 0 = 3
5 x
5
√
x
3
- 1
2 x
√
x
.
(b) y 0 =
(1+cos x ) sec
2
x +tan x sin x
(1+cos x )
2
.
(c) y 0 =
√
21 + πx
√
21+ π - 1 cos x - x
√
21+ π sin x .
(d) y
0
= -
cos( x - sin x )(1 - cos x )
sin
2
( x - sin x )
.
(e) y 0 =
- π sin(cos
√
sin( πx )) cos(cos
√
sin( πx )) sin
√
sin( πx ) cos( πx )
√
sin( πx )
.
Question 2. Find an equation for the tangent line and normal line to the curve at the given
point.
(a) y = (1 + 2 x )
2
, (1 , 9) .
(b) y =
√
x
x + 1
, (4 , 0 . 4) .
(c) x
2
+ 2 xy - y
2
+ x = 2 , (1 , 2) .
Solution.
1

2 MATH 155A FALL 13
The slope of the normal line is -
1
12
, hence the normal equation is
y - 9 = -
1
12
( x - 1) .
(b) Similar to (a); the two equations are y - 0 . 4 = - 0 . 03( x - 4) and y - 0 . 4 =
100
3
( x - 4).
(c) Use implicit differentiation to get
y
0
=
- 2 x - 2 y - 1
2 x - 2 y
.