Prelim 2 Practice problems and solutions
Problem 1 (15 points) Find the derivative of the following functions:
√
(a) (5 points) f (x) = 3 3x4 + 4x + 7
Answer: Let u = 3x4 + 4x + 7, then du
= 12x3 + 4. By chain rule,
dt
f 0 (x) =
du
d 1/3
d 1/3 du
1
1
u =
u
= u−2/3
= (3x4 +4x+7)−2/3 (12x3 +4)
dx
du
dx
3
dx
3
ln(x2 +1)+x
ln(7x+2)+3x
2
(b) (5 points) g(x) =
Answer: let u(x) = ln(x + 1) + x, v(x) = ln(7x + 2) + 3x.
u0 (x) = (ln(x2 + 1))0 + 1, v 0 (x) = (ln(7x + 2))0 + 3
By letting u = x2 + 1,
du
dx
(ln(x2 + 1))0 =
= 2x. By chain rule,
d
du
1 du
2x
ln(u)
=
= 2
du
dx
u dx
x +1
Similarly,
(ln(7x + 2))0 =
7
7x + 2
Hence,
u0 (x) =
2x
7
+ 1, v 0 (x) =
+3
+1
7x + 2
x2
Then we have,
u0 (x)v(x) − u(x)v 0 (x)
[v(x)]2
7
( 22x + 1)(ln(7x + 2) + 3x) − (ln(x2 + 1) + x)( 7x+2
+ 3)
= x +1
(ln(7x + 2) + 3x)2
g 0 (x) =
(c) (5 points) h(x) = 2e3x+1
Answer: Let u = 3x + 1, du
= 3. By chain rule,
dx
h0 (x) =
d u
d u du
du
2e =
2e
= 2eu
= 2e3x+1 (3) = 6e3x+1
dx
du
dx
dx
Problem 2 (20 points) Find the derivative of the following functions:
1
(a) (5 points) f (x) = cos(3x − 2) + sin(4x2 )
= 3, by chain rule
Answer: Let u = 3x − 2, then du
dx
d
d
du
cos(3x − 2) =
cos(u)
= − sin(u)(3) = −3 sin(3x − 2)
dx
du
dx
Similarly,
d
sin(4x2 ) = 8x cos(4x2 )
dx
Hence,
f 0 (x) = −3 sin(3x − 2) + 8x cos(4x2 )
(b) (5 points) g(x) = 3x4 + 5x3 +
√
x+1+
√ 1
3x−1
Answer: g(x) = 3x4 + 5x3 + (x + 1)1/2 + (3x − 1)−1/2 . By chain
rule,
1
1
((x + 1)1/2 )0 = (x + 1)−1/2 (x + 1)0 = (x + 1)−1/2
2
2
Similarly,
1
1
3
((3x−1)−1/2 )0 = − (3x−1)−3/2 (3x−1)0 = − (3x−1)−3/2 (3) = − (3x−1)−3/2
2
2
2
Hence,
3
1
g 0 (x) = 12x3 + 15x2 + (x + 1)−1/2 − (3x − 1)−3/2
2
2
(c) (5 points) h(x) = xex
Answer: By product rule,
h0 (x) = (1)ex + xex = ex + xex
(d) (5 points) s(x) =
√ x
x2 +1
Answer: By quotient rule,
√
√
(x)0 ( x2 + 1) − (x)( x2 + 1)0
0
√
s (x) =
( x2 + 1)2
By chain rule,
√
1
1
( x2 + 1)0 = ((x2 +1)1/2 )0 = (x2 +1)−1/2 (x2 +1)0 = (x2 +1)−1/2 (2x) = x(x2 +1)−1/2
2
2
2
Hence,
√
x2 + 1 − x(x(x2 + 1)−1/2 )
x2 + 1
√
x2 + 1 − x2 (x2 + 1)−1/2
=
x2 + 1
√
2
x2 + 1 − √xx2 +1
=
x2 + 1
0
s (x) =
(x2 +1)−x2
(x2 +1)1/2
x2 + 1
=
=
(x2
1
+ 1)3/2
Problem 3 (20 points)
2
Given f (x) = x3 e−x .
(a) Find f 0 (x).
Answer: By product rule,
2
2
f 0 (x) = (x3 )0 e−x + x3 (e−x )0
Let u = −x2 ,
du
dx
= −2x. By chain rule,
d u du
2
e
= eu (−2x) = −2xe−x
du dx
2
(e−x )0 =
Hence,
2
2
2
2
f 0 (x) = 3x2 e−x + x3 (−2xe−x ) = 3x2 e−x − 2x4 e−x
(b) Find the critical points of f (x).
Answer:
f 0 (x) = 0
2
(3x2 − 2x4 )e−x = 0
3x2 − 2x4 = 0
x2 (3 − 2x2 ) = 0
(1)
3
2
We get the third equality from the second one because e−x is never
equal to zero.
Now, we are left with two possibilities: x2 = 0 or 3 − 2x2 = 0.
The first one gives x = 0, the second one gives x2 = 32 , which
q
q
3
gives x =
, − 32 , that is x ≈ 1.22or − 1.22. And f (0) =
2
0, f (1.22) ≈ 0.41, f (−1.22) ≈ −0.41. So we have got three critical points (0, 0), (1.22, 0.41), (−1.22, −0.41).
(c) By using either first derivative test or second derivative test,
test whether the critical points are maximum or minimum or inflection.
By first derivative test, f 0 (−0.1) > 0, f 0 (0.1) > 0, so (0, 0) is inflection point. f 0 (1.2) > 0, f 0 (1.3) < 0, so (1.22, 0.41) is a maximum
point. f 0 (−1.3) < 0, f 0 (−1.2) > 0, so (−1.22, −0.41) is minimum
point.
Problem 4 (15 points)
Given f (x) = x ln x, for x > 0.
(a) Find f 0 (x).
By product rule,
1
f 0 (x) = (1) ln x + x( ) = ln x + 1
x
(b) Find the critical points of f (x).
f 0 (x)
ln x + 1
ln x
x
=
=
=
=
0
0
−1
e−1
f (e−1 ) = e−1 ln(e−1 ) = −e−1 .
So (e−1 , −e−1 ) is a critical point.
(c) By using either first derivative test or second derivative test,
test whether the critical points are maximum or minimum or inflection.
4
e−1 ≈ 0.37. By first derivative test, f 0 (0.35) < 0, f 0 (0.39) > 0 so
it is a minimum point.
OR, you can use second derivative test, f 00 (x) = x1 . f 00 (0.37) > 0
so it is a minimum point.
Problem 5 (20 points)
Suppose the cost in dollars of manufacturing q items is given by
C(q) = 2000q + 3500
and the demand equation is given by
p
q = 15000 − 1.5p
(a) Find an expression for the revenue R(q) in terms of q. (Revenue is the total amount of money the businessman can make by
selling q items at the price p dollars).
p
q =
15000 − 1.5p
2
q = 15000 − 1.5p
1.5p = 15000 − q 2
q2
p = 10000 −
1.5
Revenue, R(q) = pq = (10000 −
q2
)q
1.5
= 10000q −
q3
.
1.5
(b) Find an expression for the profit P (q) in terms of q.
Profit=Revenue-Cost, that is P (q) = R(q) − C(q). That is
P (q) = 10000q −
q3
q3
− (2000q + 3500) = 8000q −
− 3500
1.5
1.5
(c) Find the marginal profit, P 0 (q).
P 0 (q) = 8000 − 2q 2
(d) How many items should he sell to maximize the profit? How
much can he make by then?
5
P 0 (q) = 0
8000 − 2q 2 = 0
q 2 = 4000
So q ≈ 63.25, −63.25 But we reject the negative value since number
of item must be positive. P (63.25) ≈ 333810.
By first derivative test, P 0 (63) > 0, P 0 (64) < 0 so the point
(63.25, 333810) is a maximum point. But the number of items that
should be sold must be an integer, it means that it should be either
63 or 64 but not 63.25. P (63) ≈ 333802, P (64) ≈ 333737. Hence
the number of items that should be sold is 63. And the maximum
profit is 333,802 dollars.
Problem 6 (20 points)
Given the relation between x and y as follows:
x+y =5
Given an expression x2 + 9y 2 − 45, find the values of x and y so
that the value of this expression is minimized or maximized.
Answer : We write x = 5 − y for the relation x + y = 5. Plug
x = 5 − y into the expression x2 + 9y 2 − 45 to get
(5 − y)2 + 9y 2 − 45 = (25 − 10y + y 2 ) + 9y 2 − 45 = 10y 2 − 10y − 20
Call it F (y).
F 0 (y) = 20y − 10. Set F 0 (x) = 0 to get 20y − 10 = 0 and hence
y = 0.5. By first derivative test, F 0 (0.4) < 0, F 0 (0.6) > 0, so at
y = 0.5, it is a minimum point. F (0.5) = −22.5. When y = 0.5,
x = 5 − 0.5 = 4.5.
Hence, when y = 0.5, x = 4.5, the value of the expression is
minimized, and the value is -22.5.
Problem 7 (20 points) Look at Quiz 4.
Problem 8 Do at least some of the homework problems in Chapter
6.1, 6.2. A few of them, like the last problem in 6.2, are hard and I
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don’t expect to see those really hard problems in prelim. But please
make sure you know how to do at least a couple of them besides the
hard ones.
Problem 9 (20 points)
Given that f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 1
f 0 (1) = 3, f 0 (2) = 4, f 0 (3) = 1, f 0 (4) = 2
g(1) = 1, g(2) = 2, g(3) = 3, g(4) = 4
g 0 (1) = 4, g 0 (2) = 3, g 0 (3) = 2, g 0 (4) = 1
(a) (5 points) Find the derivative of ln(f (x)) at x = 1.
By chain rule, (ln(f (x)))0 =
1
f 0 (x).
f (x)
At x = 1, it becomes
1 0
1
3
f (1) = 3 =
f (1)
2
2
(b) (5 points) Find the derivative of f (g(x)) at x = 1.
By chain rule, (f (g(x)))0 = f 0 (g(x))g 0 (x). At x = 1, it becomes
f 0 (g(1))g 0 (1) = f 0 (1)4 = (3)(4) = 12
(c) (5 points) Find the derivative of
f (x)
g(x)
at x = 1.
By quotient rule, at x = 1, it becomes
f 0 (1)g(1) − f (1)g 0 (1)
(3)(1) − (2)(4)
=
= −5
[g(1)]2
(1)2
(d) (5 points) Find the derivative of [f (x)]2 at x = 1.
By chain rule, ([f (x)]2 )0 = 2f (x)f 0 (x). At x = 1, it becomes
2f (1)f 0 (1) = 2(2)(3) = 12
Problem 10 (20 points) See page 3 of the practice problem set
that was uploaded onto the course webpage
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Problem 11 To test your understanding of the things written
above, do problems 35, 36 in chapter 5.4.
Problem 12 Do problem 3 on page 1 of the practice problem set
uploaded onto the course webpage.
Problem 13 Do question 26 in chapter 5.4 in HW8.
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