Study Guide for Test 2
Selected Solutions
Part 1
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Part 1
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1. State the chain rule for taking derivatives. Include all hypotheses. -see
page 160 of text.
2. State the product rule for taking derivatives. Include all hypotheses see page 152 of the text.
3. State the quotient rule for taking derivatives. Include all hypotheses
see page 154 of the text.
4. Find the derivatives of the following functions:
(a) f (x) = sin(x).
Solution: f 0 (x) = cos(x).
(b) f (x) = x2 cos(x)
Solution: Using the product rule with u = x2 and v = cos x, we
have u0 = 2x and v 0 = − sin x. Now the product rule:
f 0 (x) = u0 v + uv 0
= 2x cos x + x2 (− sin x)
= 2x cos x − x2 sin x.
(c) g(x) = sin((cos(x))2 )
Solution: This is a nested chain rule problem. If we take
y = sin(u)
u = v2
v = cos(x)
1
dy
= cos(u)
du
du
= 2v
dv
dv
= − sin(x).
dx
Putting this together
dy du dv
dy
=
·
·
dx
du dv dx
= cos(u) · 2v · (− sin(x))
= − cos(v 2 ) · 2v · (sin(x))
= − cos((cos(x))2 ) · 2 cos(x) · sin(x).
(d) h(x) = (x2 + 1)5 (x3 + 2x)8
Solution: This is a product rule followed by two chain rules.
The last operation is a product so we start with the product rule,
letting
u = (x2 + 1)5
v = (x3 + 2x)8
dv
dx
du
dx
= 5(x2 + 1)4 (2x)
= 8(x3 + 2x)7 (3x2 + 2).
Using the product rule we have
du
dv
v+u
dx
dx
2
= 5(x + 1)4 (2x)(x3 + 2x)8 + (x2 + 1)5 8(x3 + 2x)7 (3x2 + 2)
= (x2 + 1)4 (x3 + 2x)7 10x(x3 + 2x) + 8(x2 + 1)(3x2 + 2) .
h0 (x) =
t−1
t+1
Solution: This is a quotient rule with u = t − 1, u0 = 1, v = t + 1,
adn v 0 = 1. This gives
(e) g(t) =
g 0 (t) =
1 · (t + 1) − (t − 1) · 1
2
u0 v − uv 0
=
=
.
2
2
v
(t + 1)
(t + 1)2
x2 (3x3 + 2)2
(f) h(x) =
ex
Solution: OOPS! This problem should have read
h(x) =
x2 (3x3 + 2)2
sin(x)
The last operation is division so this is done first with the quotient
rule with u = x2 (3x3 + 2)2 and v = sin(x). While it is easy
2
to calculate that v 0 = cos(x), finding u0 takes more effort. Let
u = w · z where w = x2 and z = (3x3 + 2)2 . Then w0 = 2x while
we have to do a chain rule to get z 0 = 2(3x3 + 2)(9x2 ). Using the
product rule, we now get
u0 = w0 z + wz 0 = 2x(3x3 + 2)2 + x2 (2)(3x3 + 2)(9x2 ).
Now that we have both u0 and v 0 we can use the quotient rule to
find
u0 v − uv 0
v2
(2x(3x3 + 2)2 + x2 (2)(3x3 + 2)(9x2 )) sin(x) − (x2 (3x3 + 2)2 ) (cos(x))
=
(sin(x))2
h0 (x) =
(g) f (x) = (2x(x2 − 3x))6
Solution: The sneaky easier way to do this is to simplify f (x) at
the start by multiplying the two terms inside the parentheses to
get
f (x) = (2x3 − 6x2 )6 .
We now use a chain rule, letting f (x) = y and
y = u6
u = 2x3 − 6x2
du
dx
dy
du
= 6u5
= 6x2 − 12x.
By the chain rule, we then have
f 0 (x) = 6(2x3 − 6x2 )5 (6x2 − 12x).
5. Compute
dy
dx
using the chain rule
(a) y = u3/2 and u = 4x + 2
du
dy
Solution: We have
= (3/2)u1/2 and
= 4. So
du
dx
dy
dy du
=
·
= (3/2)u1/2 · 4 = (3/2)(4x + 1)1/2 · 4.
dx
du dx
Simplifying this yields
dy
= 6(4x + 1)1/2 .
dx
3
(b) y = u(u + 1)5 and u = x2 + x
Solution: By the product rule
dy
= (u+1)5 +u·5(u+1)4 = (u+1)4 (u+1+5u) = (u+1)4 (6u+1).
du
We also have
du
= 2x + 1.
dx
Putting these together, we have
dy
= (u+1)4 (6u+1)(2x+1) = (x2 +x+1)4 (6(x2 +x)+1)(2x+1).
dx
6. Find f 0 (1) if f (x) = (−x2 + 8)4 (tan(x) + 1)4 .
Solution: Again we have a product rule, where f (x) = uv with u =
(−x2 + 8)4 and v = (tan(x) + 1)4 . We need to compute u0 and v 0 . For
u0 , we note that this is a chain rule with
du
= 4w3
dw
dw
= −2x so
dx
u = w4
w = −x2 + 8
u0 =
du
du dw
=
·
dx
dw dx
= 4w3 · (−2x) = −8x(−x2 + 8)3 .
and
dv
= 4z 3
dz
dz
= sec2 (x) so
dx
v = z4
z = tan(x) + 1
v0 =
dv
dv dz
=
·
dx
dz dx
= 4z 3 · (sec2 (x)) = 4x(tan(x) + 1)3 sec2 (x).
Now by the product rule:
f 0 (x) = u0 v + uv 0
= −8x(−x2 + 8)3 · (tan(x) + 1)4 + (−x2 + 8)4 · 4x(tan(x) + 1)3 sec2 (x)
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Part II
Calculator and Study Sheet Allowed
7. True or false
(a) The product of two functions with positive derivatives must have
a positive derivative.
Solution: False. From the product rule, (f (x)g(x))0 = f 0 (x)g(x)+
f (x)g 0 (x). If both of the functions were negative (even with positive derivative) then their product would have a negative derivative. (Consider f (x) = 1 − 1/x and g(x) = 1 − 1/x on (1, ∞).
Both have positive derivative, but their product 1−2/x+1/x2 has
derivative 2x−2 − 2x−3 , which has negative derivative for x > 1.)
(b) The composition of two functions with positive derivatives must
have a positive derivative.
Solution: True (but be careful, in this question there is the assumption that the derivative is positive for all values of x. In this
case, the chain rule tells us that
(f ◦ g)0 (x) = f 0 (g(x)) · g 0 (x)
and both terms in the product are positive, so the product is
positive.
(c) If f 0 (3) > 0, g(2) = 3 and (f ◦ g)0 (2) < 0 then g 0 (2) < 0.
Solution: True: By the chain rule
0 > (f ◦ g)0 (2) = f 0 (g(2)) · g 0 (2) = f 0 (3) · g 0 (2),
Since f 0 (3) is positive, we can deduce that g 0 (2) must be negative.
(If the question was as in the original version with g 0 (3) < 0, then
we could have concluded nothing about g 0 (3), and consequently
the statement would have been False.)
dy
at the point (1, 2).
8. Suppose x3 y 2 + y 3 = 12. Find dx
dy
Solution: We need to find
and for this we use implicit differentidx
ation. Differentiating both sides of the equation with respect to x we
get
dy
dy
3x2 y 2 + x3 (2y ) + 3y 2
= 0.
dx
dx
5
Given that we are looking at the point (1, 2), we may assume x = 1
and y = 2 so that we have
12 + 4
Solving for
dy
gives
dx
dy
dy
+ 12
= 0.
dx
dx
dy
12
−3
=− =
dx
16
4
at the point (1, 2).
If we were further asked to write the equation of the tangent line at
that point, it would be given by
y−2=
−3
(x − 1).
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9. Suppose we have the functions f (x) and g(x), both of which are differentiable. Suppose further that we have the following chart:
f (x)
g(x)
f 0 (x)
g 0 (x)
0
5
−1
2
0
1 2 3 4 5 6 7
2 −3 1 2 6 3 7
1 0 4 7 −2 2 6
6 5 4 6 7 1 0
2 −2 5 0 6 7 1
(a) If h(x) = f (x)g(x), find h0 (6).
Solution: By the product rule h0 (x) = f 0 (x)g(x) + f (x)g 0 (x), so
at x = 6 we have
h0 (6) = f 0 (6)g(6) + f (6)g 0 (6)
= 1 · 2 + 3 · 7 = 23.
(x)
, find h0 (3).
(b) If h(x) = fg(x)
Solution: By the quotient rule,
h0 (x) =
f 0 (x)g(x) − f (x)g 0 (x)
,
(g(x))2
6
so at x = 3 we have
f 0 (3)g(3) − f (3)g 0 (3)
(g(3))2
4·4−1·5
=
(4)2
11
=
.
16
h0 (3) =
(c) If h(x) = f (g(x)), find h0 (4).
Solution: By the chain rule h0 (x) = f 0 (g(x))g 0 (x). At x = 4 we
have:
h0 (4) =
=
=
=
f 0 (g(4)) · g 0 (4)
f 0 (7) · 0
0·0
0.
10. The mean arterial pressure in a human is given by the formula
P =O·R
where P denotes the mean arterial pressure, O denotes the cardiac
output rate, and R denotes the systemic vascular resistance..
(a) Find the rate of change of P with respect to time t in terms of O,
and dR
.
R, dO
dt
dt
Solution: We take derivatives with respect to t of the equation
above, and by the product rule we have:
dO
dR
dP
=
·R+O·
.
dt
dt
dt
(b) A typical value for R is 1400 dyne-sec/cm5 and for O is 87.5 cm3 /sec.
If when a typical person starts exercising, their cardiac output (O)
increases at a rate of 20 cm3 /sec/sec, what rate does the systemic
vascular resistance need to change at to keep the rate of change
of the pressure P to less than 1000 dyne/cm2 /sec. Explain your
reasoning.
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Solution: The problem above tells us that we can use R = 1400
dO
and O = 87.5 and
= 20. Thus we have
dt
dP
dt
We want
dP
dt
dR
dt
dR
= 28000 + 87.5 .
dt
= 20 · 1400 + 87.5
< 1000, so we are left with the inequality
28000 + 87.5
dR
< 1000.
dt
Solving this we have
87.5
dR
< −27000,
dt
or
dR
−27000
<
= −308.57 . . . ( dyne-sec/cm5 )/sec.
dt
87.5
By the way, the units in this problem are an example of the need
to manage conversions. Blood pressure is usually measured in
millimeters of mercury, Cardiac output is usually measured in
Liters/minute, and systemic vascular resistance is typically measured in dyne-seconds/cm5 . Of course, none of these units works
directly with any of the others.
11. Look at the two graphs below these graphs show the distance (in meters) versus time (in seconds) of a pitched ball (at around 95 mph) from
home plate, and the angle (in degrees) that the batter makes from his
head and eyes to the ball (assuming his body is square to the plate)
to watch the ball when it is some distance (in meters) away from the
plate. Thus when the ball is 2 meters from the plate, his head and eyes
are around an 80 degree angle, and when the ball is over the plate, it is
a 0 degree angle. Use the two graphs to approximate the instantaneous
rate of change of the angle when the ball is 2 meters over the plate and
when it is directly over the plate. Explain your reasoning and relate it
to the chain rule.
8
distance vs time
meters
15
10
5
0.1
0.2
0.3
0.4
0.5
seconds
angle vs. distance
degrees
80
60
40
20
0.5
1.0
1.5
2.0
2.5
meters
-20
Solution: The idea here is that we have two different functions given
to us by their graphs. Namely, the function of the distance x based
on the time t from the ball being thrown (we will call this x(t)), and
the function of the angle A based on the distance x of the ball to
home plate (we will call this A(x). Composing these functions, we
get a new function A(t) = A(x(t)). The problem is asking us for the
instantaneous rate of change of the angle with respect to time, so we are
dA
. Since we have A given to us as a composition,
looking at finding
dt
this involves using the chain rule, which in this case is
dA
dA dx
=
· .
dt
dx dt
dA
dx
Now, we want to know how to find
and
from the graphs at the
dx
dt
when the ball is 2 and 0 meters from the plate. Since the derivative is
given by the slope of the tangent line, we need to find the slopes of the
two curves at x = 2 and x = 0 (remembering that in the first graph, x
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is on the vertical axis as time (t) is on the horizontal axis). The graph
of x(t) is a straight line, so the slope (of the tangent) will be the same
whether we are at x = 0 or x = 2. Judging from the graph, this slope
is roughly
18 − 2
dx
=
= −40m/s.
dt
0 − .4
On the other graph, x is on the x-axis, and at x = 2, the slope is
roughly (eyeballing the tangent line)
80 − 70
= 4 deg /m
2.5 − 0
and at x = 2 roughly (again eyeballing the tangent line)
20 − 0
= 200 deg /m
.1 − 0
Thus at x = 2 when the ball is two meters from the plate, the instantaneous rate of change in the angle is
dA
dA
dx
|x=2 =
|x=2 · |t=.4 = 4(−40) = −160 deg /s,
dt
dx
dt
and at x = 0 when the ball is two meters from the plate, the instantaneous rate of change in the angle is
dA
dx
dA
|x=0 =
|x=0 · |t=.44 = 200(−40) = −8000 deg /s,
dt
dx
dt
Note: I think in class, I substituted one of the values incorrectly
(maybe 10 rather than −40?). The above is correct.
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