Dr. Sageev, Math 121 S
Second Quiz
1. Solve the equation ln(ln(3x)) = 0.
2. Find the point on the graph of y = (1 + x2 )ex , where the tangent line is horizontal.
3
3. The volume of a spherical tumor is given by V = πx6 where x is the diameter of the tumor.
A physician estimates that the diameter is growing at the rate of 0.4 mm/day, at a time
when the diameter was 10 mm. How fast is the volume of the tumor changing at that time?
4. Differentiate the following functions:
√
√
c)y = 2x · x x
a) y = e2x b) y = ln 10x+1
x−1
5. a) Find the relative extreme point of f (x) =
or a relative max.
ln x+1
,
x
and b) determine if it is a relative min
6. a) Determine that the point (0, e) is on the graph of exy = ln y, and b) find
dy
dx
at that point.
7. A sample of radioactive material has decay constant 0.25. Time is measured in days. What
will the sample size be when the sample is decreasing at the rate of 2 grams/day?
8. One hundred shares of tech stock were purchased on January 2, 1990 for $1200 and sold
on January 2, 1998 for $1250. What rate of interest compounded continuously did this
investment earn?
9. Solve the indefinite integral:
Z 1
3
3x
2e + 2 − + 1 dx
x
x
1
Dr. Sageev, Math 121 S
Second Quiz
Solutions
1. We need to undo the logs:
eln(ln(3x)) = e0
ln(3x) = 1
eln(3x) = e1
3x = e
x = e/3
2. “Where the tangent line is horizontal” means that y 0 = 0. So take the derivative using the
product rule, set equal to 0, solve for x:
y 0 = (2x)ex + (1 + x2 )ex
0 = ex (2x + 1 + x2 )
0 = (x + 1)2 ,
so x = −1 and plug back in to find y = 2/e.
3. This is a related rates problem. We want to find dV
, and after we interpret the problem,
dt
we are given that dx
=
0.4
when
x
=
10.
Since
V
is
given in terms of x and we want the
dt
derivative in terms of t, so we get
dV dx
dV
=
dt
dx dt 3πx2
=
(0.4)
6
= π102 (0.4)/2
= 20π.
1
4.a. If we simplify y to (e2x ) 2 = ex , then we just have y 0 = ex .
4.b. Using log rules, we can simplify this a lot:
y = ln |x + 1| − (x − 1) ln |10|,
so
y0 =
1
− (1) ln(10).
x+1
2
Dr. Sageev, Math 121 S
Second Quiz
4.c. Use logarithmic differentiation:
√ ln y = ln 2x · x x
√
= ln(2x ) + ln(x x )
√
= x ln(2) + x ln x
√ 1
1 0
1 1
y = ln(2) + ( x− 2 )(ln x) + ( x)( )
y
2
x
√
√ 1
1
1
y 0 = (2x · x x )(ln(2) + ( x− 2 )(ln x) + ( x)( ))
2
x
5.a. “Relative extreme point” means derivative! Use the quotient rule.
x( x1 ) − (1)(ln x + 1)
x2
1 − ln x − 1
=
x2
− ln x
0=
x2
0 = ln x,
f 0 (x) =
so x = 1, and f (1) = 1, so (1, 1) is an extreme point.
5.b. One way is to use the second derivative test. Then
x2 (− x1 ) − 2x(− ln x)
x4
−1 − 0
f 00 (1) =
1
= −1,
f 00 (x) =
so f (x) is concave down at x = 1, so (1, 1) is a relative maximum.
6.a. We just need to check that (0, e) satisfies the equation, so plug it in:
e0 = ln e
and both equal 1, so this works.
6.b. Use implicit differentiation:
3
Dr. Sageev, Math 121 S
Second Quiz
exy = ln y
1
(xy)0 exy = y 0
y
y0
((1)y + x(y 0 ))exy =
y
y0
(e)e0 =
e
2
e = y0.
(plug in (0, e))
7. We are given that the decay constant r = −0.25 and the sample is decreasing at some rate,
so P 0 = −2. We have the equation
P 0 = rP,
and we want to know what P is. Then plug these in and solve for P and we get P = 8
grams.
8. The initial value is P0 = 1200. At t = 8 years, we have
P (8) = 1200er(8)
1250 = 1200er(8)
1250
= er(8)
1200
1250
ln
= 8r
1200
ln 1250
1200
= r.
8
9. Recall your antiderivative rules, and we find that
Z 1
3
2
3x
2e + 2 − + 1 dx = e3x − x−1 − 3 ln |x| + x + C
x
x
3
4