BC Calculus 1
Sample Quiz
Show all appropriate work clearly for full credit.
NO CALCULATORS
1.
Name:
For each function shown below, find its first derivative. DO NOT simplify.
 e( x  2 x ) 
2
f ( x)  ln 
  x  2 x  ln(cosh( x))
cosh(
x
)


2
a.
f   x  2x  2 
sinh( x)
cosh( x)
b. g  x   tan 1 ( x3  3)  sec2 x
g   x   tan 1 ( x3  3)  2(sec x)(sec( x) tan( x)  sec 2 ( x)
2.
2x
1  ( x 2  3) 2
Find f ''  x  if f ( x)  sin  x 2 
f ( x)  2 x cos  x 2  
f ( x)  2 x  2 x sin  x 2   2cos  x 2 
3.
Find values of p and q such that the line y  px  5 is tangent to the graph of
q
 x   x3   1 at the point where x = 1.
x
q
p     2  2  p  q  3
I.
1
II.
At x = 1, the tangent line and f have the same value  p  5  2  q  p  q  7
Therefore p  5 and q  2
3a. Suppose h( x)   f ( g ( x))  . Given the information about , g and their derivatives provided in
the table, fill in the missing information.
2
x
1
2
4
f  x
3
–1
5
f  x
–1
8
6
g ( x)
4
2
1
g ( x)
1
4
3
h( x )
25
1

h( x )
60
-64

Work space:
3b. Given the information above, suppose that k ( x)  f ( g ( x)) where k is one-to-one and
d
 k 1 ( x )  at x = 5.
differentiable. Evaluate the
dx
d
1
 k 1 ( x)  
dx 
k (k 1 ( x))
Now, k (1)  5  k 1 (5)  1 , and
k ( x)  f ( g ( x)) g( x)  k (k 1 (5))  k (1)  f ( g (1)) g(1)  f (4) g(1)  6 1 .
d
1
1
 k 1 ( x)  x 5 
Therefore,

1
dx
k (k (5)) 6
ln( x)  1
. Determine the exact value of x for which this function has a stationary
x
point, then determine whether k has a local max, a local min, or neither at this point. Be sure
to justify.
Let k  x  
4.
1
x   (ln( x)  1
 ln( x)
k x  x

2
x
x2
ln( x)  1
has a stationary point at x = 1.
x
 ln( x)
 ln( x)
Since k   x  
 0 when x  (0,1), and k   x  
 0 when x  (1, ) ,
2
x
x2
it follows that k has a local maximum at x = 1.
k   x   0   ln( x)  0  x  1 . So k  x  
5.
Consider the curve defined by the equation 3 y3  12 x2 y  16 x3  16 .
a.
dy
.
dx
24 xy  48 x 2
Use implicit differentiation to determine


9 y 2  y  12 x 2 y  24 xy  48x 2  0  y 
b.
9 y 2  12 x 2

8 xy  16 x 2
3 y 2  4 x2
Find any point(s) on the curve where the tangent line to the curve is horizontal.
8xy 16 x2  0  8x( y  2 x)  0  x  0 or y  2 x
Looking at the curve 3 y3  12 x2 y  16 x3  16
x  0  3 y3  16  y  3
16
,and
3
y  2 x  3(2 x)3 12 x2 (2 x)  16 x3  16  16 x3  16  x  1

16 
So, there are horizontal tangents at  0, 3
 and (1,2)
3 

Concepts:
6. Show that if x2 y 2  xy  42 , then
dy  y
.

dx
x
d 2 2
d
dy
dy
x y  xy   (42)  2 xy 2  2 x 2 y  1 y  x  0

dx
dx
dx
dx
dy
  2 x 2 y  x   2 xy 2  y
dx
dy  y (2 xy  1)


dx
x(2 xy  1)
dy  y


dx
x
7. A function y  f ( x) is said to satisfy the Grand L. Liu Condition if it satisfies the formula
f ( x)
f ( x)  2 . Find all functions that satisfy the Grand L. Liu Condition.
x
We want a function that when you differentiate, you get an x 2 in the derivative; something like 
1
.
x
1
1
f ( x)
 1
But if f ( x)   , f ( x)  2  2 . So we need to have g    , where g doesn’t change when we
x
x
x
 x
differentiate. Hmmmm, do we know such a function? How about e x ?
Let’s try:
1
1


1
f ( x)
x
f ( x)  e , f ( x)  e x  2  2 .
x
x
Note: If C is any real constant, f ( x)  Ce

1
x
works as well.
d
f ( x)
[ Ln  f ( x) ] 
,
dx
f ( x)
2
d
1
[ Ln  f ( x) ]  2 . So, let f ( x)  e( something whose derivative is 1/ x )
Then , we want
dx
x
Alternately: Noticing