# SOLUTIONS TO END OF CHAPTER EXERCISES

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```					SOLUTIONS TO SELECTED END OF CHAPTER EXERCISES

1.1   a.   y = 3x3 - 2x2 + 9x + 4

dy
 9x2  4x  9
dx

b.   y = 13x0

Since any base raised to the zero power is equal to one then

y = 13x 0 = 13(1) = 13

dy
and       = 0.
dx

c.   y = (4x 2 + 2) (3x 3 + 12x)

dy
= (4x 2 + 2) (9x 2 + 12) + (3x 3 + 12x) (8x)
dx
= (36x 4 + 48x 2 + 18x 2 + 24) + (24x4 + 96x 2)

= 60x4 + 162x 2 + 24

100
d.   y=       2
= 100x- 2
x

dy
= - 200x -3
dx

2 x 2  3x
e.   y=
6x  2

dy (6 x  2)(4 x  3)  (2 x 2  3x )(6)
=
dx              ( 6 x  2) 2

24 x 2  18 x  8 x  6  12 x 2  18 x
=
36 x 2  24 x  4

12 x 2  8 x  6
=
36 x 2  24 x  4

6x 2  4 x  3
=
18 x 2  12 x  2
1.3   a.      y = 4x 4 + 3x 3 + 2x 2 + x + 1

f' (x) = 16x 3 + 9x 2 + 4x + 1

f'' (x) = 48x 2 + 18x + 4

f''' (x) = 96x + 18

b.      y = (2x 3 + x 2) (x 2 - 5x)

f' (x) = (2x 3 + x 2) (2x - 5) + (x 2 - 5x) (6x 2 + 2x)

= 4x 4 - 10x 3 + 2x 3 - 5x 2 + 6x 4 + 2x 3 - 30x 3 - 10x 2

= 10x 4 - 36x 3 - 15 x2

f'' (x) = 40x 3 - 108x2 - 30x

f''' (x) = 120x 2 - 216x - 30

1
c.      y=     3
 x 3
x

f' (x) = -3x -4

f'' (x) = 12x -5

f''' (x) = -60x -6

1 3
1.5   TC =     q  2q 2  6q  10
3

dTC
MC =        q 2  4q  6
dq

dMC
First Order Condition:            2q  4  0
dq

2q = 4

q* = 2

d 2 MC
Second Order Condition:         2  0 indicates that MC is minimized when q = 2
dq 2
units. The minimum value of MC is found by substituting q* = 2 into the MC function as
follows.
MC* = (q*)2 - 4q* + 6 = 22 - 4(2) + 6 = \$2.

1.7   a.      y = 8x 2z 2 - 2x 3z 3
y
 16 xz 2  6 x 2 z 3
x

y
 16 x 2 z  6 x 3 z 2
z

dy y y dz
b.            
dx x z dx

dy                                           dz
= (16xz 2 - 6x 2z 3) + (16x 2z - 6x 3z 2)
dx                                           dx

dy y dx y
   
dz x dz z

dy                         dx
 (16 xz 2  6 x 2 z 3 )  (16 x 2 z  6 x 3 z 2 )
dz                         dz

```
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