1- Find dy/dx if:
−1 (x)
y = ex+tan
1
dy
−1
= ex+tan (x) 1 +
dx
1 + x2
Using the Chain Rule, we have
2-
ln(xy ) +
1
y
−1 (x)
= ex+tan
2 + x2
1 + x2
= sec(x)
Solution. First we simplify this to get:
y ln x + y −1 = sec(x).
y(ln x)0 + (ln x)(y)0 − y −2 (y)0
dy
1 dy
1
− 2
y · + ln x ·
x
dx y dx
1
dy
(ln x − 2 )
dx
y
dy
dx
= sec x tan x
= sec x tan x
= sec x tan x −
sec x tan x −
=
ln x − y12
y
x
y
x
2
esin y + ln( xy ) = 1
3-
Solution. First we simplify this to get:
esin y + 2 ln x − ln y = 1.
Now we dierentiate with respect to x :
(esin y )(sin y)0 +
2 1 dy
−
= 0
x y dx
esin y cos y ·
dy sin y
1
−2
(e
cos y − ) =
dx
y
x
4-
y=
dy 2 1 dy
+ −
= 0
dx x y dx
−2
dy
=
sin
y
dx
x(e
cos y − y1 )
2x
1 + 2x
Applying the Quotient Rule yields,
2x ln(2) · (1 + 2x ) −
√ 2x · 2x ln(2)
2x ln(2)
dy
=
=
dx
(1 + 2x )2
(1 + 2x )2
5-
2
xey = tan−1 (ex ) √ y
Using implicit differentiation we have
2
2
ey + xey 2yy 0 =
ex
√ y 0 so,
1 + e2x
1
ex
2
(xe 2y + 1)y =
√ ey =⇒ y 0 =
2x
1+e
2xyey2 + 1
y2
0
ex
2
√ ey
2x
1+e
6-
7-
Find the derivative of y = log2 (5x3 ).
Solution
dy
dx
=
=
=
=
=
=
=
d
log2 (5x3 )
dx
!
d ln 5x3
dx ln 2
d
1
· (ln 5 + 3 ln x)
ln 2 dx
!
1 d
d
ln 5 + (3 ln x)
ln 2 dx
dx
!
1
d
0+3·
ln x
ln 2
dx
1
1
·3·
ln 2
x
3
x ln 2
Change of Base Formula
Constant Multiple Rule and Log Properties (3) & (6)
Term by Term Differentiation
Derivative of Constant & Constant Multiple Rule
Derivative of ln
8-
Find the area of the largest rectangle that can be inscribed in the
first quadrant of the unit circle if one side of the rectangle lies along the x-axis
and another lies along the y-axis.
(Hint: The equation of a unit circle centered at the origin is x2 + y 2 = 1.)
y
1
y‡
1 - x2
Hx,yL
We wish to maximize the area A of the rectangle given y =
√
1 √ x2 .
p
A = xy = x 1 √ x2
p
√2x
A0 = x · √
+ 1 √ x2
2 1 √ x2
p
√x2
√2x2 + 1
=√
+ 1 √ x2 = √
1 √ x2
1 √ x2
1
Find the critical points where A0 = 0 given x > 0.
√
√2x2 + 1 = 0 ⇒ x2 = 1/2 ⇒ x = 1/ 2
Confirm that there is a maximum
value at the critical point using the first
derivative test. Since A0 12 > 0 and A0 45 < 0, there is a maximum value
√
√
at x = 1/ 2, y = 1/ 2. The area of the largest inscribed rectangle is
√ 2
1/ 2 = 1/2 .
9- A 20-ft ladder is leaning against a wall when its base starts to
slide away. At the moment when the angle between the ladder and the ground
is π/3 radians, the top of the ladder is sliding down the wall at a rate of 1/4
ft/sec. How fast is the base of the ladder moving away from the wall then?
20 ft
y
Solution:
Let x represent the distance from the base of the ladder to the wall and y
represent the distance from the top of the ladder to the ground. When
√ the
angle θ between the ladder and√the ground is π/3 radians, then y = 10 3 ft
and x = 10 ft because sin θ = 3/2 and cos θ = 1/2. We wish to find dx/dt
given dy/dt = √1/4 ft/sec. Use the Pythagorean Theorem.
x2 + y 2 = 202
dx
dy
2x
+ 2y
=0
dt
dt
dx
dy
x
+y
=0
dt dt
√
1
dx
=0
10
+ (10 3) √
dt
4
√
dx
3
=
ft/sec
dt
4
Θ
x
x
10- An open-top box is to be made by cutting congruent squares of side length x from the corners of a 20- by 25-inch sheet of tin and bending up the sides (Figure 4.38). How large
should the squares be to make the box hold as much as possible? What is the resulting
maximum volume?
SOLUTION
Model The height of the box is x, and the other two dimensions are 20 2x and
25 2x. Thus, the volume of the box is
x
Vx x20 2x25 2x.
x
x
x
20"
x
x
x
x
25"
(a)
.
Solve Graphically Because 2x cannot exceed 20, we have 0 x 10. Figure 4.39
suggests that the maximum value of V is about 820.53 and occurs at x 3.68.
Confirm Analytically Expanding, we obtain Vx 4x 3 90x 2 500x. The first
derivative of V is
Vx 12x 2 180x 500.
The two solutions of the quadratic equation Vx 0 are
180 18
0 2
48
500
c1 3.68
24
25 2x
and
x
180 18
0 2
500
48
c2 11.32.
24
20 2x
(b)
Only c1 is in the domain 0, 10 of V. The values of V at this one critical point and the
two endpoints are
Critical point value:
Vc1 820.53
Endpoint values:
V0 0,
V10 0.
Interpret Cutout squares that are about 3.68 in. on a side give the maximum volume,
about 820.53 in3.
11-
A particle is moving along the x-axis with position function
xt 2t 3 14t 2 22t 5,
t 0.
Find the velocity and acceleration,
SOLUTION
Solve Analytically
The velocity is
vt xt 6t 2 28t 22 2t 13t 11,
and the acceleration is
at vt xt 12t 28 43t 7.
Notice that the first derivative (v x) is zero when t 1 and t 11/3.
The acceleration a(t) 12t 28 has a single zero at t 7/3.
12-
A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6
mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar
that the distance between them and the car is increasing at 20 mph. If the cruiser is moving
at 60 mph at the instant of measurement, what is the speed of the car?
SOLUTION
We carry out the steps of the strategy.
Let x be the distance of the speeding car from the intersection, let y be the distance
of the police cruiser from the intersection, and let z be the distance between the car
and the cruiser. Distances x and z are increasing, but distance y is decreasing; so
dy/dt is negative.
We seek: dx/dt
We know: dz/dt 20 mph and dy/dt 60 mph
A sketch (Figure 4.56) shows that x, y, and z form three sides of a right triangle. We
need to relate those three variables, so we use the Pythagorean Theorem:
x2 y2 z2
Differentiating implicitly with respect to t, we get
dx
dy
dz
dx
dy
dz
2x 2y 2z , which reduces to x y
z
.
dt
dt
dt
dt
dt
dt
2 y2 ):
We now substitute the numerical values for x, y, dz/dt, dy/dt, and z (which equals x
dx
(0.8)
(0.6)(60) (0.8)2 (0.6
)2 (20)
dt
dx
(0.8) 36 (1)(20)
dt
dx
70
dt
At the moment in question, the car’s speed is 70 mph.
z
y
x
Figure 4.56 A sketch showing the variables in Example 3. We know dy/dt and
dz/dt, and we seek dx/dt. The variables x,
y, and z are related by the Pythagorean
Theorem: x2 y2 z2.
5 ft
13-
Water runs into a conical tank at the rate of 9 ft 3min. The tank stands point down and
has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the
water is 6 ft deep?
r
h
Filling a Conical Tank
10 ft
Figure 4.57 In Example 4, the cone of
water is increasing in volume inside the
reservoir. We know dV/dt and we seek
dh/dt. Similar triangles enable us to relate
V directly to h.
SOLUTION 1
We carry out the steps of the strategy. Figure 4.57 shows a partially filled conical tank.
The tank itself does not change over time; what we are interested in is the changing
cone of water inside the tank. Let V be the volume, r the radius, and h the height of the
cone of water.
We seek: dh/dt
We know: dV/dt = 9 ft3/min
We need to relate V and h. The volume of the cone of water is V 13
pr 2h, but this
formula also involves the variable r, whose rate of change is not given. We need to
either find dr/dt (see Solution 2) or eliminate r from the equation, which we can do by
using the similar triangles in Figure 4.57 to relate r and h:
h
r
5
, or simply r .
2
h
10
Therefore,
h 1
2 h .
1
h
V p 3
2
2
p
3
Differentiate with respect to t:
dV
p
dh
p dh
• 3h2 h2 .
dt
12
dt
4
dt
Let h 6 and dV/dt 9; then solve for dh/dt:
p
dh
9 (6)2 4
dt
dh
1
0.32
dt
p
At the moment in question, the water level is rising at 0.32 ft/min.
SOLUTION 2
The similar triangle relationship
h
dr
1 dh
r also implies that 2
dt
2 dt
and that r 3 when h 6. So, we could have left all three variables in the formula
V 13
r 2h and proceeded as follows:
dV
1
dr
dh
p 2r hr2 dt
3
dt
dt
1
1 dh
dh
p 2r hr2 3
2 dt
dt
1
1 dh
dh
9 p 2(3) (6)(3)2 3
2 dt
dt
dh
9 9p dt
dh
1
dt
p
This is obviously more complicated than the one-variable approach. In general, it is computationally easier to simplify expressions as much as possible before you differentiate.
14-