(Boyd&Scott)1
A long time ago, in a galaxy far, far away there were the……
Table of Contents:
Do Not Stray From the Path Young Padawan
Real World Applicability………………………………………………………………….2
Analytical Example……..…………………………………………………………………….3
Multiple Choice Question………………………………………………………………….3-4
Conceptual Example…………………………………………………………………………4-5
Graphing Calculator Problem..………………………………………………………….6-7
AP Level Free Response………………………………………………………………..7-8
Mathematician Who Contributed………………………………………………………..9
Try Your Hand At These (Excercises)……………………………………………..10-13
Works Cited…………………………………………………………………………………….14
* All images and references used in this project are property of Lucas Arts Ltd.
(Boyd&Scott)2
Hello, and welcome to our chapter on the first and second derivative and
its applications. We should probably start off by explaining, just what are
the applications of the first and second derivatives?
APPLICATIONS OF THE DERIVATIVE
The derivative is probably the most applicable concept in calculus (yay!). It is
most widely applied to the field of physics. The first derivative of a position
function finds the instantaneous velocity of the object being “tracked” and the
second derivative finds the instantaneous acceleration. They can be applied to
many scenarios such as throwing a rock, punting a Ewok, driving a tie fighter
into a space colony, and piloting an X-wing through the narrow passages of the
death star. The derivative is also applicable to the business world. The
derivative allows one to find points of optimization (maximums or minimums)
of functions, such as ones that show profit; this could be used to show the
maximum gain of Jabba the Hutt.
Here are the disciplines your Jedi master (master Groden or Miller) has taught
you the ways of. Please, meditate on them for a moment and feel free to come
back here at any point. They are indispensible.
Power Rule:
𝒅
𝒅𝒙
Quotient Rule:
[𝒖𝒏 ] = 𝒏𝒖𝒏−𝟏
𝒅
𝒖
[ ]=
𝒅𝒙 𝒗
𝒗(𝒖′ )−𝒖(𝒗′ )
𝒗𝟐
Chain Rule:
𝒅
𝒅𝒙
[𝒇(𝒈(𝒙))] = 𝒇′ (𝒈(𝒙)) ∙ 𝒈′ (𝒙)
Product Rule:
𝒅
𝒅𝒙
[𝒖𝒗] = 𝒖(𝒗′ ) + 𝒗(𝒖′ )
And now that you are well informed, let us begin our journey. The
journey of the derivative, may the calculus be with you…..
Analytical example: simple power rule differentiation.
(Boyd&Scott)3
Your mission: discover the second derivative of 𝑓(𝑥 ) = 𝑥 3
𝑓(𝑥 ) = 𝑥 3
Function
𝑓 ′ (𝑥 ) = 3𝑥 2
Use power rule
𝑓 ′′ (𝑥 ) = 6𝑥
Use power rule
Explanation: Even padawans can complete this journey of power,
remember; if 𝑓(𝑥 ) = 𝑥 𝑛 , 𝑓 ′ (𝑥 ) = 𝑛𝑥 𝑛−1 . In simpler terms, all you need to
do is “pull down (With your force pull)” the exponent, multiply it by the
function you are differentiating, and then subtract 1 from that exponent.
Analytical mission 2: the constant rule.
𝑓(𝑥 ) = 1138
Function as a constant
𝑓 ′ (𝑥 ) = 0
Use the constant rule
Conceptual behind analytical:
The slope of a horizontal line is 0, so the function that describes the slope
of the original function (Known as the derivative) has to equal 0.
Remember, the derivative of a number is ALWAYS 0, end of story.
The mission of many choices: the cyclical derivatives
If 𝑓(𝑥 ) = cos 𝑥 what does 𝑓 ′′ (𝑥 ) =
a) cos 𝑥
b) sin 𝑥
c) 0
d) − sin 𝑥
e) − cos 𝑥
a) cos 𝑥
So close to correct, you be. Change of
signs you have forgot!
b) sin 𝑥
You are a sith.
(Boyd&Scott)4
c) 0
cos(x) is cyclical, never equals 0
d) -sin 𝑥
YOU ONLY TOOK ONE DERIVITIVE!!!
Always continue on, young Padawan.
e) -𝐜𝐨𝐬 𝒙
Don’t give in to the dark side
derivatives can get tricky when applying them to trigonometric
functions,because you must remember both the sign changes and the function
changes. By remembering the two basic trigonometric rules, the derivative of
𝐜𝐨𝐬(𝒙) = −𝐬𝐢𝐧(𝒙), the derivative of 𝐬𝐢𝐧 𝒙 = 𝐜𝐨𝐬 𝒙you can take as many derivatives
as you like, because these functions are cyclical.
Who needs calculus?
Light sabers solve all
problems.
Test your mind now, we will (conceptual example): The derivative of
a graph over an interval.
The graph of 𝑓(𝑥) is shown below determine the derivative of the function when 𝑥 =
0.1, 0.3, 𝑎𝑛𝑑 0.6.
Answer: 𝒇′(𝟎. 𝟏) = 𝟓, 𝒇′(𝟎. 𝟑) = 𝟎 , 𝒇′ (𝟎. 𝟔) = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅.
Explanation: Derivatives of points on lines are particularly simple to
find because the slope at a specific point (aka the derivative) is
equal to the slope over the interval (0, 0.2) (the linear part on the
(Boyd&Scott)5
function). Since slope is equal to
find𝒇′ (𝟎. 𝟏) =
𝒔𝟐 −𝒔𝟏
𝒕𝟐 −𝒕𝟏
, you would use the equation to
𝟏.𝟎𝟎−𝟎
.𝟐𝟎−𝟎
, 𝒇′ (𝟎. 𝟏) = 𝟓
The slope of a horizontal line is always zero, so 𝒇′ (𝟎. 𝟑) = 𝟎, that
is all young apprentice.
Differentiate a “cusp” (a sharp turn) cannot you. Find you not
that sneaky derivative at 𝒕 = 𝟎. 𝟔𝟎.
Uuuuuuuuuuuurrrrrr Arrrrrggggg
Die Wookie!
“Does your brain feel like a X-wing fighter jumping to
hyper speed? You need to take a break master, and enjoy the
following picture.”
(Boyd&Scott)6
“Feeling better? Good, now continue on with your training, master.”
Weapons training: graphing calculators at the ready!
If 𝒇(𝒙) = 𝒂𝒓𝒄𝒕𝒂𝒏(𝟓𝟎𝟎𝒙), what is 𝒇′ (𝟏𝟔𝟐. 𝟓𝟕𝟏) = (GC required)
𝒇′ (𝟏𝟔𝟐. 𝟓𝟕𝟏) = 𝟕. 𝟓𝟔𝟓 × 𝟏𝟎−𝟖
How did the Jedi masters come to such a conclusion?
Let the force guide you…and follow these steps.
Enter the function tan−1 (500x) into 𝑌1 of your graphing calculator. Then
hit MATH (not with your force lightning, you little sith) and scroll down to
nDeriv(. Go to vars, press the option for 𝑌1 , then press X, and type in the
number 162.571 [Make sure you are in radians], close parentheses and
(Boyd&Scott)7
hit enter. You have discovered the meaning of nothingness.
Congratulations! You can differentiate with your calculator! Join the lazy
club (otherwise known as the light side), we have pie (𝜋!).
Luke Needs Your Help: Find His Light saber!
During the epic battle between Luke Skywalker and Darth Vader, Luke’s
light saber falls from a 72 meter high platform. The position function for
the plummeting light saber is 𝑠(𝑡) = −3𝑡 2 − 8𝑡 + 72 when 0 ≤ 𝑡 ≤ 3.841
a. Find the velocity of the light saber when it has fallen exactly 20
meters.
b. Find the time when the light saber is falling the fastest. If there is
none, explain why.
a. 𝟐𝟎 = −𝟑𝒕𝟐 − 𝟖𝒕 + 𝟕𝟐
𝒕 = 𝟑. 𝟎𝟑𝟖
first, you need to find the time when
the light saber has fallen 20 meters,
So set 𝒔(𝒕) equal to 20
𝒔′ (𝒕) = −𝟔𝒕 − 𝟖
then, take the derivative to find a
function that models the velocity
of this given position function
(Boyd&Scott)8
𝒔′ (𝟑. 𝟎𝟑𝟖) = −𝟐𝟔. 𝟐𝟐𝟖
last, plug in the time found when
𝒔(𝒕) = 𝟐𝟎 into the derivative to
find the velocity
b. 𝒔′′ (𝒕) = −𝟔
𝒔′′ (𝒕) = 𝟎 When 𝒕 = 𝟎, If 0 is plugged into the original equation,
𝒔′ (𝒕), to find the velocity, one would find it equal to -8. However,
since acceleration becomes more negative, as time goes on.
Since you are looking for an absolute minimum of velocity, you
have to check endpoints. 0 was already checked to give -8, but
when you plug in 3.841, you would receive 𝒔′ (𝟑. 𝟖𝟒𝟏) = −𝟑𝟏. 𝟎𝟒𝟔.
Therefore giving the absolute minimum over the interval.
a. 1- find where t=20
1- Correct derivative of 𝒔′ (𝒕)
1- Correct answer
b. 1 - Find zeroes of 𝒔′′ (𝒕)
1 - Justify end points
1 - Correct answer
The almighty one!! (Mathematician who contributed to derivatives).
One might question who would come up with such a powerful tool. That
genius went by the name Gottfried Leibniz (1646 – 1716). He was responsible
for inventing calculus at the same time but independent of Newton. Leibniz
was born in Leipzig, Germany; he is known as “the last universal genius”. He
was a fantastic philosopher, geologist, physicist, historian, and (most
importantly to us) a mathematician. He was guided by the leading
(Boyd&Scott)9
mathematician in Europe, Christian Huygens, but was surprisingly
completely self taught in mathematics. He studied the relationship between
difference sequences and sums, and then an infinitesimal version that helped
him discover the essential background of calculus. He also discovered the
difference quotient (𝐥𝐢𝐦 𝒇(𝒙) =
𝒉→𝟎
𝒇(𝒙+𝒉)− 𝒇(𝒙)
𝒉
= 𝒇′ (𝒙)).
Leibniz’s main studies as a philosopher resulted in the style of his work in
calculus; he was more focused on creating a general symbolic language than
finding specific results. This is why his notations, and not Newton’s, are used
in today’s calculus.
Our guidance has come to an end, you have finally become a Jedi master. As a
master, you must overcome the challenges that face you, whether they are
difficult or not. Go on your path and achieve greatness.
Let us begin with some basic training. Solve the problems, do whatever it
takes to accomplish the task at hand. Now go!
Find 𝒚′ of the given functions
𝒚 = 𝟐𝒙𝟐 + 𝟒𝒙 + 𝟏𝟐𝟔
𝒚=
𝟏
𝟐𝒙𝟐
𝒚 = −𝟕
Find 𝒚′′ of the given functions
𝟒
𝒚 = √𝒙
𝒚 = − 𝐜𝐨𝐬 𝒙 + 𝒙𝟖
Very impressed, the master are with you. But can you complete these more
difficult problems?
Use implicit differentiation to find the first derivative
𝒚𝟐 = 𝟒𝒚𝟑 + 𝟐𝒙𝟐 + 𝟏𝟔𝒙
( 𝒙 + 𝟏) 𝟐 ( 𝒚 − 𝟏) 𝟐
𝟏𝟐 =
+
𝟐
𝟒
𝒅𝒚
𝒅𝒙
of the given functions
(Boyd&Scott)10
𝒚𝟑 − 𝟕𝒙 = 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒚
Find 𝒚′ of the given functions
𝒚 = (𝒙𝟐 + 𝟐)(𝒙 − 𝟏)
𝒙𝟑 + 𝟔𝒙𝟐 − 𝟖
𝒚=
𝒙
𝒚 = 𝐭𝐚𝐧 𝒙 𝐬𝐢𝐧 𝒙
Find 𝒚′′ of the given functions
𝒚 = (𝐜𝐨𝐬 𝒙𝟑 )𝟐
𝒚=
𝐬𝐢𝐧 𝒙
𝒆𝒙𝟓𝝅
𝒚 = (𝒙𝟐 + 𝟏𝟐)(𝟔𝒙𝟑 − 𝟐𝒙 + 𝟏)𝟐
𝒚=
𝟏
√𝒙𝟓 𝐬𝐢𝐧 𝒙
If you are able to complete these last 5 difficult problems. You have truly
mastered the mystery of the derivative. We are much pleased.
Find 𝒚′ of the given functions
𝒚 = 𝐬𝐢𝐧(𝐜𝐨𝐬(𝐭𝐚𝐧(𝐬𝐢𝐧(𝟔𝒙))))
(𝟗𝒙𝟐 − 𝟖𝒙 + 𝟕)
𝒚=
× 𝐜𝐬𝐜 𝒙𝟐
𝐬𝐢𝐧 𝒙𝟑
3
√𝑥 2 − 4𝑥 + 9
𝑦=
sin(cos(𝑥 3 ))
Find 𝑦 ′′ of the given function
𝒚 = 𝐬𝐞𝐜(𝝅√𝒙)
Solve using implicit differentiation
𝟐𝟕 =
𝟒
𝟏
𝟐
𝐬𝐢𝐧
𝒙
+
𝐬𝐢𝐧
𝒚
𝐜𝐨𝐬
𝒚𝟑
𝒙
We shall now practice some AP level multiple choice questions. So prepared
you shall be, for the AP test should you take that challenge.
(Boyd&Scott)11
Find an equation to the tangent line to the graph of 𝟐𝒙𝟑 − 𝟓𝒙 +
a.)
b.)
c.)
d.)
e.)
𝟑
𝒙𝟐
𝒘𝒉𝒆𝒏 𝒙 = 𝟐
𝒚 = 𝟏𝟎. 𝟐𝒙 − 𝟓
𝒚 = 𝟏𝟗. 𝟐𝟓(𝒙 − 𝟐)
𝒚 − 𝟔. 𝟕𝟓 = 𝟏𝟗. 𝟐𝟓(𝒙 − 𝟐)
𝒚 + 𝟔. 𝟕𝟓 = 𝟏𝟗. 𝟐𝟓(𝒙 − 𝟐)
𝒚 + 𝟐 = 𝟏𝟗. 𝟐𝟓(𝒙 − 𝟔. 𝟕𝟓)
How many places does the function 𝒇(𝒙) = 𝟑𝒙𝟐 − 𝟓𝒙𝟑 + 𝟐 change it’s concavity?
a.)
b.)
c.)
d.)
e.)
One
Two
Three
Four
Five
If the function given by 𝑓(𝑥) = 6𝑥 3 − 4𝑥 2 − 3𝑥 + 2, when does 𝑓(𝑥)have a relative
minimum?
A.)
0.687
B.) -0.243
C.) -0.707
D.) -0.687
E.) 0.707
Find 𝑓′′′(𝑥) given 𝑓 ′ (𝑥) =
A.)
B.)
C.)
D.)
E.)
5𝑥 3 −2
𝑥3
𝑥 3 +2
𝑥3
30𝑥 3 −6
𝑥4
−6
𝑥4
3𝑥
𝑥 3 −1
𝑥2
(Boyd&Scott)12
5
Given 𝑓(𝑥) = 5𝑒 ln( √𝑥
4 +9𝑥 2 −2)
, find 𝑓 ′ (𝑥)
5
A.) 5√𝑥 4 + 9𝑥 2 − 2
B.)
C.)
4𝑥 3 +18𝑥
5
√(𝑥 4 +9𝑥 2 −2)4
1
5
√(𝑥 4 +9𝑥 2 −2)4
D.) 20𝑥 3 + 90𝑥
5
E.) (4𝑥 3 + 18𝑥)√(𝑥 4 + 9𝑥 2 − 2)4
(Boyd&Scott)13
AP Free Response Problem: Let the force guide you.
1.) The Aluminum falcon has to escape from deadly hyperspace, filled with
asteroids flying at incredible speeds. The Aluminum falcon’s position is monitored
by the function 𝑆(𝑡) when 0 ≤ 𝑡 ≤ 9.000
t (minutes) 0
S(t)
(Lightyears) 0
1
2
3
1.031
-1.782 1.290
4
5
6
0.859
-0.516 0.371
7
8
9
1.523
3.981
12
a.) What is the average velocity of the aluminum falcon travels between 𝑡 =
1 𝑎𝑛𝑑 𝑡 = 2? Indicate units of measure.
b.)
9
Use the information in the table to approximate ∫3 𝑣(𝑡). What does this
mean in the context of the problem?
c.) From time 𝑡 = 0 𝑡𝑜 𝑡 = 9, what is the minimum amount of relative extrema that
this function must have?
d.)
If after time 𝑡 = 10, the ship’s path is monitored by the function 𝑆(𝑡) =
3𝑡 2 − 82𝑡 + 9 What is the acceleration at time 𝑡 = 24.5?
(Boyd&Scott)14
http://apcentral.collegeboard.com/apc/members/courses/teacher
s_corner/31217.html
http://www.physics.brocku.ca/courses/1P21_Crandles/Kinematics
/
http://www.tutorcircle.com/application-of-differentiationt17Jp.html#
http://sofia.nmsu.edu/~history/book/leibniz.pdf
http://plato.stanford.edu/entries/leibniz/
http://apcentral.collegeboard.com/apc/public/repository/ap2012_calcu
lusbc_frq.pdf