Download 1. Calculate the derivatives of the following: (a) g(x) = √ 3x Solution

1. Calculate the derivatives of the following:
√
(a) g(x) = 5 3x
Solution:
1
3
g 0 (x) = (3x)−4/5 · 3 =
5
5(3x)4/5
(b) h(x) = x3 + 3x2 + 2x + 1
Solution:
h0 (x) = 3x2 + 6x + 2
(c) f (x) =
x2 + 1
x−2
Solution:
(x − 2)(2x) − (x2 + 1)(1)
(x − 2)2
2x2 − 4x − x2 − 1
=
(x − 2)2
x2 − 4x − 1
=
(x − 2)2
f 0 (x) =
2. Calculate the derivatives of the following:
−x3
(a) q(x) =
.
10
Solution:
q 0 (x) =
(b) y =
√
−3 2
x.
10
7x2 − 3x
Solution:
(c) g(x) = 3x4 sin x.
dy
1
14x − 3
= (7x2 − 3x)−1/2 (14x − 3) = √
dx
2
2 7x2 − 3x
Solution:
g 0 (x) = 3x4 cos x + 12x3 sin x
= 3x3 (x cos x + 4 sin x)
3. Find the exact value of the limit
√
cos x −
limπ
x→ 4
x − π4
2
2
Solution: Since this limit is the derivative
√ of the function cos x at the point π/4,
the value of the limit is − sin (π/4) = − 2/2.
Solution: The hard way: Let θ = x − π4 . When x →
x = π4 + θ.
√
cos x −
limπ
x→ 4
x − π4
2
2
π
,
4
θ →
π
4
−
π
4
= 0, and
cos x − cos π4
x→ 4
x − π4
cos( π4 + θ) − cos π4
= lim
θ→0
θ
π
cos( 4 ) cos(θ) − sin( π4 ) sin(θ) − cos π4
= lim
θ→0
θ
cos( π4 ) cos(θ) − cos π4
sin( π4 ) sin(θ)
= lim
− lim
θ→0
θ→0
θ
θ
π π cos(θ) − 1
sin(θ)
= cos
lim
− sin
lim
4 θ→0
θ 4 θ→0 θ
π
π
= cos
· 0 − sin
·1
4 4
π
= − sin
√ 4
2
=−
2
= limπ
Where does the idea for setting θ = x − π4 come from? Either from recognizing
that this is the derivative of cosine, or from remembering that we know the limit
limθ→0 cos(θ)−1
= 0, and noting that we need a single variable for the denominator.
θ
Page 2
4. Find
d2 y
dt2
of y = −4.9t2 − 31.5t + 15.35
Solution: It’s −9.8. We can view this as acceleration due to gravity where this is a
position function of a projectile.
Otherwise,
dy
dt
= −9.8t − 31.5 and so
d2 y
dt2
= −9.8.
5. A tank which has the shape of an inverted circular cone is being filled at a rate of 25
cm3 /min. If the top of the tank has a diameter of 10cm and a height of 12 cm, find the
rate at which the depth of the water is increasing when the water is 6cm deep.
Solution: First, draw a picture!
5 cm
r
12 cm
h
Using a similar triangles argument, the ratio of
5
12
= hr , so r =
5
h.
12
The equation we need is
2
1 2
1
5
V = πr h = π
h3
3
3
12
1 25 3
= ·
πh
3 144
and so its derivative with respect to time is:
dV
25 2 dh
=
πh
dt
144
dt
In this problem: dh
is the unknown,
dt
h = 6cm. Giving us:
dV
dt
= 25cm3 /min, at we’re interested when
dV
25 2 dh
=
πh
dt
144
dt
25
dh
25cm3 /min =
π (6cm)2
144
dt
25cm3 /min
dh
=
25
dt
π (6cm)2
144
4
dh
=
π
dt
Page 3
so
dh
dt
≈ 1.2732cm/min.
6. Find the equation of the tangent line to the curve f (x) = sin x − cos x at the point (π, 1).
Solution: Here we just find the slope, which will be f 0 (π), so first: f 0 (x) = cos x +
sin x, and then
f 0 (π) = cos π + sin π = −1 + 0 = −1
so the equation of the line is:
y − 1 = −1(x − π)
y =1−x+π
y =1+π−x
7. Find the derivative of y = tan x using the quotient rule.
Solution:
d
d sin x
cos x · sin x − sin x(− sin x)
tan x =
=
dx
dx cos x
(cos x)2
2
sin x + cos2 x
=
cos2 x
1
=
cos2 x
= sec2 x
8. Find
dy
dx
2
implicitly:
(a) x + 3y 3 = 4x + y
Solution:
dy
dy
=4+
dx
dx
dy
9y 2 − 1
= 4 − 2x
dx
dy
4 − 2x
= 2
dx
9y − 1
2x + 9y 2
Question 8 continues. . .
Page 4
(b) x3 + 2 = cos y +
1
x
Solution:
1
dy
− 2
dx x
1
= −3x2 − 2
x
−3x2 − x12
=
sin y
1
2
= − 3x + 2 csc y
x
3x2 = − sin y
sin y
dy
dx
dy
dx
dy
dx
if you wanted to convert to csc y.
9. Let’s approximate
√
2.
(a) Find the linearization of f (x) =
√
4 − 8x about a = 0.
Solution: We need to find f 0 (x), f (a), and f 0 (a), and plug it all into the
formula: f (x) ≈ f (a) + f 0 (a)(x − a). Applying the power and chain rules, we
have:
−4
1
f 0 (x) = (4 − 8x)−1/2 (−8) = √
2
4 − 8x
√
−4
Since a = 0, f (a) = 4 = 2, f 0 (a) = √
= −2, and so the linearization is:
4
f (x) ≈ f (a) + f 0 (a)(x − a)
= 2 + (−2)(x − 0)
= −2x + 2
(b) What value of x do you plug into f (x) so that f (x) =
Solution: If x = 1/4,
√
2?
p
√
√
4 − 8(1/4) = 4 − 2 = 2.
(c) Use your linear
√ approximation from (a) and the x value from (b) to find an approximation for 2.
Solution: Plugging x = 1/4 into f (x) ≈ −2x + 2 yields:
√
2 = f (1/4) ≈ −2(1/4) + 2 = 1.5
Question 9 continues. . .
Page 5
(d) My calculator says
√
2 ≈ 1.41421356. How close was your approximation?
Solution: Not really very close. . . . It’s actually off on the tenths digit.
10. We’ll be approximating
√
4
16.4.
(a) Find the equation of the tangent line to f (x) =
√
4
16 + 4x at the point x = 0.
Solution: First,
1
1
f 0 (x) = (16 + 4x)−3/4 (4) =
4
(16 + 4x)3/4
√
so that f 0 (0) = 1613/4 = 18 , and f (0) = 4 16 = 2, so we have:
L(x) = f (a) + f 0 (a)(x − a)
1
= 2 + (x − 0)
8
1
=2+ x
8
(b) Using
(a), call
L(x) the equation of the tangent line. Now approximate the value
p
√
4
4
of 16.4 = 16 + 4(0.1) by plugging in an appropriate value for x into L. Note:
the actual value in a calculator is 2.01238449.
Solution: Here we need x = 0.1 since
into the linearization L(x), to get:
p
√
4
16 + 4(0.1) = 4 16.4. Just plug x = 0.1
1
L(0.1) = 2 + (0.1) = 2 + 0.0125 = 2.0125
8
which is pretty close to the value given.
x
does not have an absolute maximum on the interval [−5, 3].
(x + 2)2
Why does this not violate the Extreme Value Theorem?
11. The function g(x) =
Solution: Since x = −2 is a discontinuity of g, and −2 ∈ [−5, 3], the Extreme Value
Theorem doesn’t apply.
12. Verify that the function satisfies the hypotheses of the Mean Value Theorem on the
given interval. Then find all numbers c that satisfy the conclusion of the Mean Value
Page 6
Theorem.
f (x) =
x
,
x+2
[1, 4]
Solution: Since rational functions are differentiable, hence continuous, on their domains, and f is defined on [1, 4], we have f is continuous on [1, 4] and differentiable
on (1, 4). Thus, by the mean value theorem, there exists a c in (1, 4) such that
f (4) − f (1) = f 0 (c)(4 − 1). First, let’s find the derivative:
(x + 2)(1) − x(1)
(x + 2)2
2
=
(x + 2)2
f 0 (x) =
And f (4) =
4
6
= 32 , f (1) = 13 , so we have:
2
− 13
2
3
=
(c + 2)2
3
2
1
=
2
(c + 2)
9
18 = (c + 2)2
√
± 18 − 2 = c
f 0 (c) =
and those are the values of c = −2 ±
√
√
18 = −2 ± 3 2.
13. Prove that the equation x101 + x51 + x − 1 = 0 has exactly one real root.
Solution: Consider the function f (x) = x101 + x51 + x − 1. We prove the equation
above has only one real root if we show the polynomial function f has exactly one real
zero. First, we note that it has at least one real zero, by applying the Intermediate
Value Theorem. f (0) = −1 and f (1) = 1 + 1 + 1 − 1 = 2, so there exists a real zero
in the interval (0, 1).
[Strictly increasing argument] Now, to show that there are no additional zeroes,
consider the derivative, f 0 (x) = 101x100 + 51x50 + 1. Notice that f 0 (x) > 0 for all x.
Thus, the function is increasing for all x, and as such, will never be zero again.
[Rolle’s Theorem argument] Now, to show there are no zeroes other than z1 , let us
assume that there is another zero, call it z2 . Then, since f is a polynomial (hence
differentiable and continuous on any interval), and f (z1 ) = f (z2 ) = 0. So Rolle’s
Theorem guarantees us a c in the interval (z1 , z2 ) such that f 0 (c) = 0. But f 0 (x) 6= 0
for any x, a contradiction. Thus there are no addtional zeroes.
Page 7
14. Sketch the graph of the function that satisfies the given conditions:
• f (0) = 0 , f is odd (that is, symmetric about the origin),
• f (2) = −4,
• f 0 (x) < 0 for 0 < x < 2, f 0 (x) > 0 for x > 2,
• f 00 (x) > 0 for 0 < x < 3, f 00 (x) < 0 for x > 3,
• limx→∞ f (x) = −2.
6
5
4
3
2
1
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1
−1
1
2
3
4
5
6
7
8
9 10
−2
−3
−4
−5
−6
15. Find two positive integers (counting numbers) such that the sum of the first number
and four times the second number is 1000 and the product of the numbers is as large as
possible.
Solution: Define the numbers as x and y. We are given that x + 4y = 1000 and xy
is as large as possible. Let’s write the first equation as x = 1000 − 4y and maximize
the function xy = (1000 − 4y)y = 1000y − 4y 2 = f (y):
f 0 (y) = 1000 − 8y
So the critical point is when f 0 (y) = 0 or y = 125. Since we know that our function
f (y) is a quadratic with a negative lead coefficient, we know that y = 125 will be an
absolute maximum of the function. Thus, the two numbers are 125 and 500.
Page 8
16. We want to enclose a small rectangular yard using 48 ft of fence, using the side of our
house as one side of the rectangle. Find the maximum area.
Solution: Let x be the length of one side of a rectangle, and y be the length of the
adjacent side. In this setup, since we’re making a fence with three sides, we have
48 = 2x + y or y = 48 − 2x. Since we want to find the maximum area, which we
know is xy = x(48 − 2x) = 48x − 2x2 , we need to find the maximum of the function
A(x) = 48x − 2x2 , a quadratic with negative lead coefficient, so we know the critical
point is the absolute maximum of the function.
To find the critical number, we find A0 (x) = 48 − 4x = 0 or x = 12. This means that
y = 24, and that answers the question.
Page 9
17. Find the limit:
√
lim
x→∞
3 + x6 + x
x3 + 5x + 2
Solution:
√
√
3 + x6 + x
3 + x6 + x 1/x3
lim 3
= lim 3
·1 3
x→∞ x + 5x + 2
x→∞ x + 5x + 2
/x
√
6
1
3 + x + x /x3
= lim 3
·1 3
x→∞ x + 5x + 2
/x
√
3+x6
3
lim x 5
x→∞ 1 + 2
x
√
3+x6
√
x6
+ x12
+ x23
1
+ 2
x
= lim
5
2
x→∞
1+ 2 + 3
x
r x
6
1
3+x
+ 2
6
x
x
= lim
2
5
x→∞
1+ 2 + 3
x
r x
6
3+x
1
+ 2
6
x
x
= lim
2
5
x→∞
1+ 2 + 3
x
r x
1
3
+1+ 2
6
x
x
= lim
5
2
x→∞
1+ 2 + 3
x
x
!
r
1
3
+1+ 2
lim
x→∞
x6
x
=
5
2
lim 1 + 2 + 3
x→∞
x
x
r
3
lim 6 + 1 + 0
x→∞ x
=
√ 1+0+0
1
=
1
=1
=
Page 10
18. Given the graph, find the limits:
5
4
3
2
1
−20−19−18−17−16−15−14−13−12−11−10−9 −8 −7 −6 −5 −4 −3 −2 −1
−1
1 2 3 4 5 6 7
−2
−3
−4
−5
(a) lim f (x) =
x→∞
Solution: 0
(b) lim f (x) =
x→−∞
Solution: 3
19. Given that cosh−1 x = ln(x +
√
x2 − 1), derive the formula for the derivative:
d
1
cosh−1 x = − √
dx
x2 − 1
Page 11
Solution: I haven’t covered the derivative of ln(x) yet. I am used to the “Early
Transcendentals” version of the textbook where ln(x) and ex are covered in chapter
3. Anyway, the derivative of ln(x) is x1 , so
√
d
d
cosh−1 x =
ln(x + x2 − 1)
dx
dx
√
d
1
√
(x + x2 − 1)
=
(x + x2 − 1) dx
1
d√ 2
√
=
· 1+
x −1
dx
x + x2 − 1
1
1
d 2
√
=
(x − 1)
· 1+ √
x + x2 − 1
2 x2 − 1 dx
1
1
√
· 1+ √
2x
=
x + x2 − 1
2 x2 − 1
1
1
1
√
√
=
+
· √
2x
2
2
x + x − 1 x + x − 1 2 x2 − 1
1
x
√
√
√
=
+
2
2
x + x − 1 (x + x − 1) x2 − 1
1
x
√
√
√
=
+
2
2
x + x − 1 (x + x − 1) x2 − 1
√
x2 − 1 + x
√
√
=
(x + x2 − 1) x2 − 1
1
=√
2
x −1
Error: there shouldn’t be a minus sign in the statement of the problem.
20. Find the limit
ln(x + 1)
x→0
x
lim
Solution: see # 19
21. Find the derivative and simplify as much as possible for full credit!
ln(sin x) −
Page 12
1 2
sin x
2
Solution: see # 19
1
cos x
d
ln(sin x) − sin2 x =
− sin x cos x
dx
2
sin x
cos x − sin2 x cos x
=
sin x
cos x(1 − sin2 x)
=
sin x
cos x cos2 x
=
sin x
= cot x cos2 x
22. Find the derivative of the function
√
g(x) = arctan( x)
Solution: Omitted so that this can be a bonus problem on the Take Home
23. Find the second derivative,
d2 y
dx2
and simplify as much as possible for full credit!
y = x2 ln x
Solution:
dy
1
= 2x ln x + x2
dx
x
= 2x ln x + x
d2 y
1
= 2 ln x + 2x + 1
2
dx
x
= 2 ln x + 3
24. Find the limit.
lim (cos x)1/x
x→0+
Page 13
Solution: This is an example of indeterminant type 1∞ , so we first write:
y = (cos x)1/x ≡ ln y =
1
ln cos x
x
and we’ll work with the right-hand-side and then return back to find the solution of
the original problem:
lim+
x→0
1
ln cos x
x
is an indeterminant of type ∞ · 0, so can write it as:
lim+
x→0
1
ln cos x
ln cos x = lim+
x→0
x
x
= lim+
− sin x
cos x
x→0
1
= lim+ − tan x
x→0
=0
And now we return to the original problem, in which we exponentiate the value just
found:
lim+ (cos x)1/x = e0 = 1
x→0
25. Solve for θ:
log2 2 sin2 θ = 2 log2 cos θ + 1
Page 14
Solution:
log2 2 sin2 θ
log2 2 sin2 θ
log2 2 sin2 θ
log2 2 sin2 θ
log2 2 sin2 θ − log2 2 cos2 θ
log2 tan2 θ
2 log2 tan θ
log2 tan θ
tan θ
tan θ
= 2 log2 cos θ + 1
= log2 cos2 θ + 1
= log2 cos2 θ + log2 2
= log2 2 cos2 θ
=0
=0
=0
=0
= 20
=1
π
θ=
4
and, more generally, θ =
π
4
+ nπ.
26. Find the limit
lim x2 e2x
x→−∞
Solution: This is indeterminant type 0 · ∞, so let’s write it as:
x2
x→−∞ e−2x
2x
x
1
= lim
= lim
= lim
= lim 2e2x = 0
x→−∞ −2e−2x
x→−∞ −e−2x
x→−∞ −2e−2x
x→−∞
lim x2 e2x = lim
x→−∞
27. Find the limit (if you use a shortcut, explain why it applies)
x
4
lim 1 +
x→∞
x
Solution: The answer is e4 . This compound interest formula.
28. Solve the equation for x:
Question 28 continues. . .
Page 15
(a) ex = 3
Solution:
ex = 3
ln(ex ) = ln(3)
x = ln(3)
(b) tan−1 x =
π
4
Solution:
π
4
π
tan(tan−1 x) = tan( )
4
π
x = tan( )
4
x=1
tan−1 x =
29. (Read carefully:) The derivative of F (x) is given:
F 0 (x) =
x2 (x + 2)(3x − 5)
x−8
(a) Find the intervals on which F (x) is increasing and intervals on which it is decreasing.
Solution: The critical numbers are 0, −2, 5/3, and 8, so we just need to determine the sign of the derivative in those regions between critical numbers:
Factor x2 x + 2 3x − 5 x − 8 f 0 is. . .
f is . . .
Region \
x < −2
+
negative decreasing
−2 < x < 0
+
+
positive increasing
0 < x < 5/3
+
+
positive increasing
5/3 < x < 8
+
+
+
negative decreasing
8<x
+
+
+
+
positive increasing
So the function F (x) is increasing on the intervals (−2, 5/3) ∪ (8, ∞).
(b) Using part (a), determine the x− values of the local extrema.
Solution: x = −2 is the location of a local minimum. (f (−2) is a local min.)
Local max at x = 35 , local min at x = 8.
Page 16
√
1
and x1 = 1:
x−2
(a) Find x2 of Newton’s Method.
30. Given the function f (x) =
Solution: f 0 (x) =
x−
1
1
√ +
, so
2 x (x − 2)2
√
1
x1 − 2
x2 = x1 −
1
1
√ +
2 x1 (x1 − 2)2
√
1
1−
1−2
=1−
1
1
√ +
2 1 (1 − 2)2
1+1
=1− 1
1
+ (−1)
2
2
2
=1− 1
+1
2
2
=1− 3
x1 −
2
4
=1−
3
1
=−
3
(b) Choosing x1 = 1 was convenient for a test to make the arithmetic easy, but it was a
bad choice for Newton’s Method if we actually want to find the root. Explain why.
31. Suppose that f (0) = −1, and f 0 (t) ≤ 6. What is the largest that f (10) could be?
Solution: By the Mean Value Theorem, we know that there is some c in (0, 10) such
that
f (10) − (−1)
f (10) − f (0)
f 0 (c) =
=
≤6
10 − 0
10
where the inequality is from f 0 (t) ≤ 6 for all t, so we have:
f (10) + 1
≤ 6 ⇐⇒ f (10) ≤ 59
10
so f (10) can be at most 59.
Page 17