Calculus I
AEMA 101
Assignment 5
Due: Tuesday, October 23, 2007 at 17:00
Find f ’(x) and f ’(c) in 1 and 2.
1. f  x    x 3  3  x    2  x 2  3  x  5  , c  0 2.
Find f ’ in the following:
x3  3  x  2
3. f  x  
x2  1
2 

4. f  x   x 4  1 

 x 1
5.
f  x  3 x 
6.
f t  

x 3
cos x
, c0
ex
8.
sec x
x
f  x    csc x  sin x
9.
f  x   x2  e x  2  x  cos x
7.

f  x 
f  x 
2  ex
10. f  x   2
x 1
cos t
t
Find an equation of the tangent line to the graph of f at the indicated point in the
following:
 
11. f  x    x  1   x 2  2  ,  0, 2  ,
12. f  x   sec x,  , 2 
3 
Determine the point(s) (if any) at which the function has a horizontal tangent line in each
of the following:
15. y  x 2  1
13. f  x   e x  sin x, 0  x  
1
16. y  3  x  2  cos x, 0  x  2  
2
x
17. To estimate the height of a building, a stone is dropped from the top of the
building into a pool of water at ground level. How high is the building if the
splash is seen 6.8 seconds after the stone is dropped, given that the position
function is s  t   4.9  t 2  V0  t  s0 for free falling objects.
14. y 
Find the derivative in each of the following.
18. f  x    3  x 2  1
4
20. y 
19. g  x   3  2  x
4
t  2
21. g  t  
1
2
1
t 2
2
22. y 
27. h  x   ln  x 2  3
x2

x2  9
1
23. f     sin 2  2   
4
24. y  sin(cos x)
28. y  ln x  x 2  1

x 1
x 1
29. y  ln
25. y  e x
26. y  x 2  e x  2  x  e x  2  e x
2
ANSWERS
1.
f  x    x3  3  x    2  x 2  3  x  5 ; c  0
f /  x    3  x 2  3   2  x 2  3  x  5    x 3  3  x    4  x  3 
f /  x   10  x 4  12  x3  3  x 2  18  x  15
f /  0   15
2.
cos x
, c0
ex
x
 sin x  e x  e x  cos x e   sin x  cos x 
sin x  cos x
/
f  x 


2 x
2 x
e
e
ex
1  0 cos x
f /  0  
 1
e0
f  x 
3.
3  x 2  3   x 2  1   x 3  3  x  2    2  x 

x3  3  x  2
/
f  x 
 f  x 
2
x2 1
 x2  1
f /  x 
3  x4  3  x2  3  x2  3  2  x4  6  x2  4  x
x
2
 1
2

x4  6  x2  4  x  3
x
2
 1
2
4.
2 
2  x4

4
f  x   x 4  1 

f
x

x




x 1
 x 1
8  x 3   x  1  2  x 4 1
8  x 4  8  x3  2  x 4
3
f /  x   4  x3 

4

x

2
x2  2  x  1
 x  1
f /  x 
f /  x 
4  x5  8  x 4  4  x3  6  x 4  8  x3 4  x5  2  x 4  4  x3

x2  2  x  1
x2  2  x  1
2  x3   2  x 2  x  2 
 x  1
2
2
5.
f  x  3 x 

1
5
1
 1

x  3  x3   x 2  3  x 6  3 x3



2

5 1
5
1
f /  x   x 6  x 3  6 
6
6  x 3 x2
6.
cos t
t

t  sin t  cos t
t  sin t  cos t
f / t  

2
t
t2
f t  
7.
sec x
x
x  sec x  tan x  sec x sec x   x  tan x  1
f /  x 

x2
x2
f  x 
8.
f  x    csc x  sin x
f /  x   csc x  cot x  cos x 
1 cos x
 1


 cos x  cos x   2  1
sin x sin x
 sin x 
f /  x   cos x   csc2 x  1  cos x  cot 2 x
9.
f  x   x 2  e x  2  x  cos x
f /  x    2  x  e x  x 2  e x   2   cos x  x  sin x 
10.
f  x 
f
11.
/
 x 
2  ex
x2  1
2  e x   x 2  1  2  e x  2  x
 x2  1
f  x    x  1   x 2  2  ,
2

2  e x   x 2  2  x  1
 x2  1
2
 0, 2 
y  y1  m   x  x1 
m  f /  0    x 3  x 2  2  x  2    3  x 2  2  x  2   2
/
y  2  2  x  y  2  x  2
12.
3

2  e x   x  1
x
2
 1
2
2
 
f  x   sec x,  , 2 
3 
y  y1  m   x  x1 
3
sin
x
/
m  f     sec x   sec x  tan x 
 2 2  2 3
2
cos x  1 
3
 
2
/  

2   3


y  2  2 3  x    y  2 3  x 
3
3

13.

f  x   e x  sin x, 0  x  
f /  x   e x  sin x  e x  cos x  e x   sin x  cos x 
m  f /  x   0  e x   sin x  cos x 
sin x  cos x  0  sin x   cos x  x 
 3
y f
 4

e

3

4
 3 , 2  e
 4
2


3
4
 sin
3
4
3
3
2
e 4 
4
2





14.
1
2
 y /  2  x 3   3  0
2
x
x
1
y  2 – This function does not have any horizontal tangent lines since its first
x
derivative is never equal to zero.
y
15.
y  x2  1  y /  2  x
m  0  y/  2  x  x  0
y  02  1  1
 0,1
16.
4
y  3  x  2  cos x, 0  x  2  
y /  x   3  2  sin x  m  y /  x   0  3  2  sin x
3  2  sin x  sin x 
3
 2 
x ,
2
3 3


3 
 
y    3   2  cos 
1
3
3
3
3
 2 
y
 3
2 
2  2  3 

 2  cos

1
  3
3
3
3

  3 

 2  2  3 

 1 and 
,
 1
 ,
3
3
3

 3

17.
18.
s  t   4.9  t 2  V0  t  s0 , the stone is dropped so V0 = 0, s0 is what we’re looking
for. After t = 6.8 seconds, the position function s(t) is equal to zero, and we have:
0  4.9  6.82  0  6.8  s0  s0  4.9  46.24  226.576 units of length.
f  x    3  x 2  1  f /  x   4   3  x 2  1  6  x  24  x   3  x 2  1
4
3
3
19.
1
1
1

g  x   3  2  x  g /  x     3  2  x  2   2   
2
3 2 x
20.
y
4
t  2
 4   t  2   y /  8   t  2  
2
2
3
8
t  2
3
21.
1
3


1
1 2
2
2  g/ t  
2  2t  

t

2

t

2


 




t2  2
2
g t  
t
t
2
 2
22.
y
y 
x2
x2  9
 y/ 
2  x  x
2
2  x  x
1
2
 9 
/
y 
/
x2  9
2
 9
1
2
1

1 2
  x     x  9 2  2  x
2
2

x3
x2  9 
x2  9
x2  9
1
3
2

2
2  x   x 2  9   x3
x3  18  x
x
x2  9
 92
5

2  x 3  18  x  x 3
3
 x2  9 2
3
23.
1
1
f     sin 2  2     f /     2  sin  2     cos  2     2
4
4
1
f /    sin  2     cos  2     sin  4   
2
24.
y  sin(cos x)  y /  cos  cos x     sin x    sin x  cos  cos x 
25.
y  e x  y /  2  x  e x
2
26.
2
y  x2  ex  2  x  ex  2  ex  y /   2  x  ex  x2  ex    2  ex  2  x  e x   2  e x
y /  2  x  ex  x2  ex  2  ex  2  x  ex  2  ex  x2  ex
27.
h  x   ln  x 2  3  h /  x  
28.

1
2 x
2 x  2
x 3
x 3
2

1
y  ln x  x 2  1  ln x   ln  x 2  1
2
1 1 1
1
x
x2 1  x2
2  x2 1
y/    2  2  x   2


x 2 x 1
x x 1
x3  x
x   x 2  1
29.
x 1 1
  ln  x  1  ln  x  1 
x 1 2 
1  1
1  1 x 1 x 1
1
y/   

 
 2
2

2  x  1 x  1 2
x 1
x 1
y  ln
6