1. Practice Problems
(1)
Solution:
d
4 sec(4x) tan(4x)
ln(sec(4x)) =
= 4 tan(4x)
dx
sec(4x)
(2) Find
dy
dx
by implicit differentiation
y = x arcsin(y)
Solution: For notational simplicity I will use y 0 =
equation with respect to x
y 0 = (1) · arcsin(y) + x ·
1
p
1 − y2
dy
.
dx
!
(y 0 )
xy 0
p
y −
= arcsin(y)
1 − y2
0
p
y0 1 − y2
xy 0
p
−p
= arcsin(y)
1 − y2
1 − y2
p
y 0 ( 1 − y 2 − x)
p
= arcsin(y)
1 − y2
p
1 − y 2 arcsin y
p
y =
1 − y2 − x
0
1
Deriving both sides of the
(3) Does
x−5
x−2
f (x) =
Have a minimum on the interval [0, 6]. If not what fails for EVT?
Solution:
lim = −∞
x→2+
So no minimum. Our function is not continuous at 2 so EVT fails.
(4) A particle is moving along the hyperbola xy = 8. As it reaches the point (4, 2) the
y-coordinate is changing at a rate of 3 cm/s.
(a) How fast is the x-coordinate of the point changing at that instant?
Solution:
dx
dy
y+ x = 0
dt
dt
⇒
xy = 8
At the point (4, 2) then
dy
dt
⇒
dx
dy x
=−
dt
dt y
= 3 so
dy x
dx
= −
= −3
dt
dt y
4
= −6 cm/s
2
(b) How fast is its distance to the origin changing at that point?
Solution:
2
2
z =x +y
at (2, 4) then z =
dz
1
=
dt
z
2
dz
1
=
dt
z
⇒
dx
dy
x +y
dt
dt
√
√
4 + 16 = 20. Hence
dx
dy
x
+y
dt
dt
1
−18
= √ (4(−6) + 2(3)) = √
cm/s
20
20
2
(5) Use linearization to approximate
√
100.5
Solution:
Use f (x) =
we see that
√
x and x = 10, tangent line is L(x) = 10 +
1
(x
20
− 100). Using this
√
1
1
100.5 = f (.5) ≈ L(.5) = 10 + (.5) = 10 +
20
40
(6) Sketch a graph of a function f that is continuous on [1, 5] and has the given properties.
Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima
at 2 and 4
3
(7) If f and g are the functions whose graphs are shown, u(x) = f (g(x)), v(x) = g(f (x)),
and w(x) = g(g(x)). Find each derivative, if it exists. If it does not exist, explain
why.
Solution:
−1
3
(−3) =
4
4
(a) u0 (1) = f 0 (g(1))g 0 (1) = f 0 (3)g 0 (1) =
(b) v 0 (1) = g 0 (f (1))f 0 (1) = g 0 (2)f 0 (1)
but g 0 (2) = dne
(c) w0 (1) = g 0 (g(1))g 0 (1) = g 0 (3)g 0 (1) =
2
(−3) = −2
3
(8) Water is leaking out of an inverted conical tank at a rate of 10,000 cm3 /min at the
same time that water is being pumped into the tank at a constant rate. The tank
has height 6m and the diameter at the top is 4m. If the water level is rising at a
rate of 20 cm/min when the height of the water is 2m, find the rate at which water
is being pumped into the tank. ( V = 13 πr2 h )
Solution:
800, 000π
+ 10, 000 cm3 /min
9
4