Math 2253 Practice Test 2 Answers Fall 2016
1. Find the derivative of each of the following functions:
(a) y  x8
(b) y  t 2  2t  3
dy
dy
 8x7
 2t  2
dx
dt
(c) g (t )  t 3
Do not forget that  is a constant, so the
answer is  times the derivative of t 3 .
g '(t )   (3t 2 )
g '(t )  3 t 2
(d) y 
1
 34 x
3x
1
1
y  x 1  3x 4
3
dy
1
1 3
  x 2  3( x 4 )
dx
3
4
dy
1
3
 2  3
dx
3x 4 x 4
2. Find the slope of the graph of the function at the given point:
(2, 18)
f ( x)  3 x 3  6
Note that the quantity that gives the slope of the graph is the derivative. In problems of
this nature you do not use the y-value. It is given just to locate the point on the curve.
f '( x)  3(3x 2 )
f '( x)  9 x 2
Slope = f '(2)  9(22 )  36
3. Find an equation of the line tangent to y  x3  x at (1, 2)
dy
dy
Note that
is used to denote the value of
when x  1
dx
dx x 1
dy
 3x 2  1
dx
dy
Slope =
 3(1)2  1  4
dx x 1
Now use the point slope form y  y1  m( x  x1 )
y  2  4( x  1)
y  2  4x  4
y  4x  2
4. Find the points at which the graph of the function has a horizontal tangent line:
1
y  x3  2 x 2  5 x
3
The tangent is horizontal at points where the derivative is zero.
dy 1
 (3x 2 )  2(2 x)  5  x 2  4 x  5
dx 3
x2  4x  5  0
( x  5)( x  1)  0
x  5, 1
Note that the question asks you to find points, so you must also find the y-coordinates.
1
125
125
125 225 100
x  5 : y  (5)3  2(5) 2  5(5)  
 50  25  
 75  


3
3
3
3
3
3
1 3
1
1
1
9
8
x  1: y  (1)  2(1) 2  5(1)   2  5   3    
3
3
3
3 3
3
8
Ans: (5, 100
3 ), (1,  3 )
5. Find the derivative of each of the following functions:
(b) f ( x) 
(a) y  x 4 2  x 2
y  x 4 (2  x 2 )
1
2
1
dy
1
1
 ( x 4 ) (2  x 2 ) 2  (2 x)  (4 x 3 )(2  x 2 ) 2
dx
2
1
dy
1
  x 5 (2  x 2 ) 2  4 x 3 (2  x 2 ) 2
dx
This simplifies to
8 x3  5 x5
x
x 3
2
f '( x) 
( x 2  3)(1)  x(2 x)
( x 2  3) 2
f '( x) 
x2  3  2 x2
( x 2  3) 2
f '( x) 
3  x2
( x 2  3) 2
2  x2
6. Find the derivative of each of the following functions
(a) f ( x)  (3x  4)100
d n
du
(u )  nu n 1
Use the generalized power rule:
dx
dx
99
f '( x)  100(3x  4)  3
f '( x)  300(3x  4)99
(b) m( x) 
x2
3  5 x3
d 2
d
( x )  x 2 ( 3  5 x3 )
dx
dx
m '( x) 
3 2
( 3  5x )
1
1
( 3  5 x3 )(2 x)  x 2  (3  5 x3 ) 2 (10 x 2 )
2

3  5 x3
( 3  5 x3 )
2 x 3  5 x3  5 x 4 (3  5 x 3 )

3  5 x3
Note that this simplifies to
1
2
6 x  5x4
3
(3  5 x3 ) 2
(c) (To test your understanding of the chain rule). You read in an article that the
1
derivative of f ( x) where f ( x)  tanh 1 x is
. Find the derivative of
1  x2
g ( x)  tanh 1 2 x3
According to the chain rule the derivative of tanh 1 u w.r.t. x is
g '( x) 
1 du

1  u 2 dx
1
 6x2
1  (2 x 3 ) 2
6 x2
1  4 x6
7. Find the derivative of each of the following functions:

(b) f ( x) 
(a) y  x 4 cos x
dy
 (4 x 3 )(cos x)  ( x 4 )( sin x)
dx
dy
 4 x 3 cos x  x 4 sin x
dx
(c) g ( x)  3x  tan x
g '( x)  3  sec x
2
x
x 3
2
f '( x) 
( x 2  3)(1)  x(2 x)
( x 2  3) 2
f '( x) 
x2  3  2 x2
( x 2  3) 2
f '( x) 
3  x2
( x 2  3) 2
(d) f ( x)  x sec x  3x tan x
You have to use the product rule on each term.
f '( x)  [1(sec x)  x(sec x tan x)]  [(3)(tan x)  (3x)(sec2 x)]
f '( x)  sec x  x sec x tan x  3tan x  3x sec2 x
8. Find the second derivative of each of the following functions.
(a) f ( x)  6 x  14 x 3
(b) h( x)  csc x
4
f '( x)  6  14( 3 x )
h '( x)   csc x cot x
f '( x)  6  42 x 4
h ''( x)  ( csc x cot x)(cot x)  (csc x)(  csc 2 x)
5
f ''( x)  42( 4 x )
f ''( x)  168 x 5
h ''( x)  csc x cot 2 x  csc3 x