MATH 263A, SUMMER I, 2005, SOLUTION OF MIDTERM 2
1. Find derivative of each function:
(a) (4 points) f (x) = cotx ¡
sinx
x¡1 .
SOLUTION: f 0 (x) = ¡csc2 x ¡ [ (cosx)(x¡1)¡(sinx)(1)
].
(x¡1)2
(b) (4 points)
f(x) = tan(x2 + 2x ¡ 1) + ex
2
+3x
.
2
SOLUTION: f 0 (x) = [sec2 (x2 + 2x ¡ 1)][2x + 2] + ex +3x (2x + 3).
2
Hence f 0 (x) = (2x + 2)sec2 (x2 + 2x ¡ 1) + (2x + 3)ex +3x :
(c) (4 points) f (x) = cscx + secx ¡ ex sinx.
SOLUTION: f 0 (x) = ¡cscxcotx + secxtanx ¡ [ex sinx + ex cosx]:
(d) (4 points)
f(x) = cot¡1 x + 3sec¡1 (2x).
1
1
p1
SOLUTION: f 0 (x) = ¡ 1+x
(2) = ¡ 1+x
2 + (3)
2 +
2x 4x2 ¡1
p 3
:
x 4x2 ¡1
(e) (4 points) f (x) = sin(cos(tan(x))).
SOLUTION: f 0 (x) = cos(cos(tanx))):(¡sin(tanx)):(sec2 x):
(f) (4 points) f (x) = ln(sin(x2 + 2x ¡ 1)).
SOLUTION: f 0 (x) =
2. Find
dy
dx
1
:[cos(x2
sin(x2 +2x¡1)
+ 2x ¡ 1)]:(2x + 2):
from each equation below.
(a) (4 points) xy = sin(xy) + xy 2 .
dy
dy
2
SOLUTION: y + x dx
= cos(xy):[y + x dy
dx ] + (y + 2xy dx ) )
dy
dy
dy
dy
y + x dx
= ycos(xy) + xcos(xy) dx
+ y 2 + 2xy dx
: Hence [x ¡ xcos(xy) ¡ 2xy] dx
=
2
y ¡ y + ycos(xy): Therefore,
dy
y 2 ¡ y + ycos(xy)
=
:
dx
x ¡ xcos(xy) ¡ 2xy
(b) (4 points) xy 3 = 4x + y 2 ¡ xy.
Hence
dy
dy
dy
dy
SOLUTION: y 3 +3xy 2 dx
= 4+2y dx
¡(y+x dx
) ) (3xy 2 ¡2y+x) dx
= 4¡y¡y 3 .
dy
4 ¡ y ¡ y3
=
:
dx
3xy 2 ¡ 2y + x
Typeset by AMS-TEX
1
2
(c) (4 points) cos(siny) = xcoty.
dy
SOLUTION: ¡sin(siny)cosy:( dx
) = coty + x:(¡csc2 y):( dy
dx ) )
dy
coty
=
:
2
dx
xcsc y ¡ sin(siny)cosy
3. (6 points) The altitude of a triangle is increasing at a rate of 2 ft/min while
its area is increasing at a rate of 5 f t2 =min: At what rate does the base of the
triangle changing when the altitude is 12 ft and the area is 120 f t2 ?
SOLUTION: Let's denote by h the altitude, and x the base of the triangle.
Then the area A is
1
A = hx:
2
We know that A; h and x are functions of the time t. Hence
dA
1 dh 1 dx
= x
+ h :
dt
2 dt
2 dt
Now for h = 12,then A = 120. Hence from the above formula we get x =
dh
Moreover at this situation, dA
dt = 5; dt = 2. Thus
(120):(2)
12
= 20.
(5):(2) ¡ (20):(2)
¡30
dx
=
=
= ¡2:5:
dt
12
12
The rate of change of the base at this situation is ¡2:5 feet per minute.
4. Find the absolute maximum and absolute minimum each function on a given
interval.
(a) (4 points) f (x) = x3 ¡ 6x2 + 9x + 4 on [0, 4].
SOLUTION: f 0 (x) = 3x2 ¡12x+9, hence the critical numbers are x = 1; x = 3.
These are in [0, 4]. Now we evaluate f (x) at x = 0; 1; 3; 4 to get the answers:
f (0) = 4; f (1) = 8; f (3) = 4; f(4) = 8. Answer: Absolute maximum is 8, and absolute
minimum is 4 (on the interval [0, 4]).
(b) (4 points) f (x) =
3x
x2 +1
on [0, 3].
SOLUTION: Notice that x2 + 1 is never zero, so f (x) is continuous everywhere
2
+1)¡(3x)(2x)
on (¡1; 1). f 0 (x) = (3)(x (x
. Set f 0 (x) = 0 ) (3)(x2 + 1) ¡ (3x)(2x) = 0 )
2 +1)2
¡3x2 + 3 = 0 ) x = 1 or ¡ 1. These are critical numbers of f (x), but ¡1 is not in
[0, 3], so we need to evaluate f(x) at 0, 1, and 3 to get the answers: f (0) = 0; f (1) =
3
9
2 ; f (3) = 10 . Hence on [0, 3] absolute minimum of f(x) is 0, absolute maximum is 1.5
.