Derivative Drill on Chain Rule:
NOTE: The first part of the worksheet only involves chain rule
using trigonometric and exponential functions. There is a second
part which incorporates inverse trig and log functions.
Derivative Drill on Chain Rule Part 1:
A. Differentiate the following:
1. f() = tan(57)
2. g() = tan (3 -
)
3. H() = tan(sin(4))
4. y = e
x
5. y = e
 sec 
6. K() =
ee
5
7. H(w) = cos(
w)-
cos(w)
8. g() = sec 4 () + sec4
9. y = esin(x) sin(ex)
1
csc  
r
10. y =
1  3r
B. Perform the following differentiations:
1. Dt (sin(at))
2. Da(sin(at))
3.
d 3y
(e cos(4  3x))
dx
4.
d 3y
(e cos(4  3x))
dy
C. Quotient Rule should only be applied when you are differentiating a function of t divided
by another function of t. For example, you should use quotient rule to differentiate a
function such as F(t) =
1  tan(3t)
. However, if your function consists of a constant
1  tan(3t)
divided by a function of t quotient rule is not only inappropriate but will likely take twice as
long as should be necessary and frequently lead to errors. To differentiate a function such
as y =
4
one should first rewrite y = 4(t + 3sin(5t)) -1 and then apply chain rule
t  3sin(5t)
to get y' = -4(t + 3sin(5t)) -2 (1+15cos(5t)). Use this technique to differentiate the following
without using quotient rule.
1. y =
5
3  e  4x
2. g(t) =
8
e
sin(x)

3. H(u) =
3
5
3


2  u 2  tan(u)   2
D. Let s(t) = 5 + 10sin(2t) be the position (cm) of a particle at time t (sec),
where 0 < t < 1.
1. For what interval(s) of time will the particle be moving backwards?
2. The critical value of a function is any place where its derivative is either zero or
undefined. Find all the critical values of the function s(t) on the interval [0, 1]. Describe
the behavior of the partical at the critical values.
3. Find the acceleration function of the particle. For what values of t in [0, 1] will the
particle's velocity be increasing?
E. The concentration C (ppm or parts per million) of a chemical in a lake is given by the
function C = C0 e
where t > 0 is time (days), C0 is the chemical's concentration at t = 0,
and k > 0 is a constant.
 kt
1. Determine the rate of change of the concentration with respect to time. Include units
with your answer.
2. Is your answer to (a) always positive or always negative? Is more chemical being added to
the lake or is the chemical being removed?
3. Larger values for the constant k indicate what about the chemical's rate of change?
F. The force of attraction between two objects of mass m and M is given by F = gmM/r 2
where g is the gravitational constant and r is the distance between the two objects. If
mass is being measured in kilograms, time in seconds, and distance r in meters, then the
units of F are in Newtons (1 N = 1 (kgm)/sec2).
1. Determine the rate of change of F with respect to distance r.
2. What are the units of dF/dr?
3. Give a physical explanation of why F'(r) is always negative.
4. Suppose the distance r between the two objects changes with time; r = r(t). How fast
will the force F change with respect to time?
5. What are the units of F'(t)?
6. Suppose r(t) = e-.004t. How fast is the force changing relative to time? What is the
sign of F'(t)? Give a physical interpretation.
G. Suppose that a population P is modeled by the differential equation
dP/dt = 3.9P - .2P1.4 where P(t) gives the # of individuals P at time t (days).
1. What are the units of P'(t)? What does P'(t) represent?
2. Use the chain rule to determine P''(t).
3. What does P''(t) represent? What are its units?
Derivative Drill on Chain Rule Part 1: Solutions:
 sec2 (3   )
2 
dy
dy
3. H'() = 4cos(4) sec2(sin(4))
4.
5.
 - e x
 - sec tan e  sec 
dx
dx
 sin w
sin  w 
5
e 5
6. DK = 5 e e
7. H'(w) =
+
2 w
2 cos  w 
A. 1. f'() = 356sec2(57)
2. g'() =
 
8. g'() = 4sec4 () tan() + 43sec(4) tan(4)
 
 
9. y'= cos(x) e sin x  sin e x  e sin x  cos e x e x
1
1
1
1
csc   cot   1  3r   3csc  
2
dy r
r
r
r
10.

2
dr
1  3r 
B. 1. Dt (sin(at)) = acos(at)
2. Da(sin(at)) = tcos(at)
d 3y
3.
(e cos(4  3x)) = 3e3y sin(4 - 3x)
dx
d 3y
4.
(e cos(4  3x)) = 3e3y cos(4  3x)
dy
C.
1.
dy
20e  4x

dx 3  e  4x 2


2. g'(x) =
 4 cos  x  e sin x 
e
sin  x 


3/ 2
5sec2 u 
3
3. H'(u) =

(2  u)2
(2  tan(u))2
D. 1. The particle moves backwards when the velocity v(t) = 20cos(2t) is negative. This
occurs on the interval (.25, .75)
2. v(t) = 0 at times t = .25, .75.
At these times the particle's motion is stopped. The
direction of motion is switching from forwards to backwards at t = .25 and from backwards
to forwards at t = .75. (Why?)
3. a(t) = -4002sin(2t). The velocity increases when a(t) > 0. This occurs for t in (.5, 1).
E. 1. C'(t) = -kC0 e
ppm/day
2. C'(t) is always negative. It is not possible to determine if the chemical is being added or
removed from the lake. However, the concentration of the chemical is decreasing. (This
could happen either by adding more water to the lake or by removing the chemical.)
3. Larger values for the constant k indicate what about the chemical's rate of change?
The larger the value of k, the more negative the rate of change of the concentration. Thus,
the larger the value of k, the more rapidly the concentration of the chemical is dropping.
 kt
F.
1. F'(r) =
2gMm
r3
2. The units are N/m.
3. As the objects move farther away from each other, the attractive force decreases; F is
a decreasing function of r and hence F'(r) < 0.
dF dF dr 2gMm dr
4.


dt dr dt
dt
r3
5. The units are N/sec.
2gMm
dF dF dr 2gMm dr
.004e 0.004t  .008gMme 0.008t
6.
=


3
3
dt dr dt
dt
0.004t
r
e




F' > 0, indicating that F is an increasing function of time.
G.
1. The units of P'(t) are #individuals/day. P'(t) represents the rate of change in the
number of individuals in the population with respect to time.
2.
d2P
2
 3.9P'(t)  2.8P0.4P'(t)
dt
3. The units of P"(t) are (#individuals)/(day)2. P"(t) is the rate of change in the
population's rate of change with respect to time.
Derivative Drill on Chain Rule Part 2:
A. Differentiate the following functions:
1. F(t) = ln(tan-1(3t))
2. g(t) = e
arcsin(t)
- 4sin-1(4et) - 4cos(ln(t))
3. y = ln(x) tan-1(x)
4. H() =
 3 
ln 

  
5. y = sin-1(ln(x)) + ln(3 - sec(x))
6. F(x) = ln(tan(x)) + tan(ln(x)) + tan(x) ln(x)
7. y = ln(ln(x)) + tan-1(tan-1(x))
B. The volume V of a sphere of radius r (in cm) is given by the formula V =
1. Compute
dV
. Provide units.
dr
4 3
r .
3
2. Now suppose that r = r(t); that is, suppose that r is changing over time. Measure t in sec.
One might for instance think of the example of a snowball that is melting. Find a formula
for
dV
. Include units.
dt
C. For each of the following pictures, write  as a function of h. Then compute
1.
2.
h
10
h


8
3. Now suppose that h is changing over time; h = h(t). Find a formula for
Derivative Drill on Chain Rule Part 2: Solutions
d
.
dt
d
.
dh
1. F(t) = ln(tan-1(3t)) F'(t) =
2. g(t) = e
g'(t) =
arcsin(t)
e arcsin t 
1  t2
1
3
tan (3t) 1  9t2
- 4sin-1(4et) - 4cos(ln(t))
16e t

1  16e2t

4 sin(ln(t))
t
dy arctan(x) ln(x)


dx
x
1  x2
3. y = ln(x) tan-1(x)
4. H() =

1
 3 
ln 

  
H'() =
3
  3
2
5. y = sin-1(ln(x)) + ln(3 - sec(x))
y' =
1
x 1  ln2 x
 sec  x  tan  x 
3  sec  x 

6. F(x) = ln(tan(x)) + tan(ln(x)) + tan(x) ln(x)
F'(x) =
2
sec2  x  sec ln x  
tan  x  

  ln  x  sec2  x  

tan  x 
x
x 

7. y = ln(ln(x)) + tan-1(tan-1(x))


1
1

y' =
xln  x   1  tan 1 x
 



 1 


2
2
  1x 


B. The volume V of a sphere of radius r (in cm) is given by the formula V =
1. Compute
dV
dV
. Provide units.
 4r2
dr
dr
4 3
r .
3
2. Now suppose that r = r(t); that is, suppose that r is changing over time. Measure t in sec.
One might for instance think of the example of a snowball that is melting. Find a formula
for
dV
dV dV dr
dr

 4r2
. Include units.
, where the units are cm3/sec
dt
dr
dt
dt
dt
C.
1.  = arcsin(h/10) (Note: This could also be done using arctangent or arccosine functions.
The inverse sine function is the easiest to work with in this example. Why?)
2.  = arctan(8/h) (Note: This could also be done using arcsine or arccosine functions. The
inverse tangent function is the easiest to work with in this example. Why?)
3.
d d dh


dt dh dt
1
 h2 
10 1  
 100 


, for the function in ex 1.




d d dh 8 
1
  8 , for the function in ex 2.

 2

 h2  64
dt dh dt
64


h
 1   2  
 h 
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