By: Florin Gjergjaj and Neomis Rodriguez
Table of Contents

 The Masterminded Authors
 Chapter 1: Limits and Continuity
 Chapter 2: Derivatives
 Chapter 3: Antiderivatives
 Chapter 4: Application Problem
The Masterminded
Authors

Florin is known for having a
difficult last name. He is a
student at the High School
for Environmental Studies.
He defeated the formidable
course known as AP
Calculus AB, and he
mastered all of the concepts.
His academic goal is to
conquer the BC course and
master the concepts just as
well. His long term goal is to
go to NYU and achieve a
medical career.
The Masterminded
Authors

Neomis is known for having a
difficult first name (if you
think it’s easy, you’re
probably pronouncing it
wrong). She is currently a
student at HSES with
several AP classes. Her
academic goals include
mastering the Spanish
language and pursuing a
scientific career.
Chapter 1: Limits and
Continuity

Limits

A limit is the calculation of f(x) as
x approaches a. The three
criteria for a limit to exist are:
1. Lim f(x) exists (right)
xa+
2. Lim f(x) exists (left)
xa3. Lim f(x) = Lim f(x) = L
xa+ xaLimits can be found three ways:
Numerically, Algebraically and
Graphically
Algebraically has three steps:
 Factor
 Simplify
 Substitute
Graphically requires tracing from
the right and the left. The f(x)
must be the same number for
both for the limit to exist.
Numerically requires a table with
x and f(x) values. It shows f(x)
as x approaches a, from both
the right and left.
Algebraically
Example:
Lim
x2
x–2
x² – x – 2
(x- 2)
(factor the denominator out and cross out like terms)
(x-2) (x+1)
1
(Simplify completely and then substitute 2 for x)
(x+1)
1
= 1/ 3
((2) + 1)
Numerically requires plugging in x values that approach a to find f(x).
Graphically requires tracing the graph from the left and right as x approaches
a (requires the same value in order for the limit to exist)
X
1.9
1.99
1.999
2
2.001
2.01
F(x)
.345
.334
.333
.333
.332
.3322
Continuity

Continuity (at a point and an open interval) requires no holes (limits), gaps,
jumps and must be unbroken.
The criteria for Continuity:
1. F(c) is defined
2. Lim f(x) exists
xc
3. Lim f(x) = f(c)
xc
If it is not unbroken, it is discontinuous. There are two types of discontinuity:
1. Removable (with a limit)
2. Non-removable (without a limit)
Continuous
Discontinuous
Chapter 2: Derivatives

Definition of the Derivative
using the Limit Process

A derivative is the slope of a function
at a point. Some derivative notations
include: f ‘(x), dy/dx, y’, d/dx f(x) or
dx [y].
The definition of tangent line with
slope m is calculated using the
formula
Lim
f(c + ∆ x) – f(c) = m
∆ x 0
∆x
The definition of a derivative using
limit process is calculated using the
formula
F ‘(x) = Lim
f (x + ∆ x) – f(x)
∆x0
∆x
The steps to solve are:
1. Substitute (x + ∆ x) for each x
variable
2. Cross out opposite signs
3. Substitute for ∆ x
 Example:
f(x) = x2 + 1 (0,1)
Lim
∆x  0
(x + ∆x)2 + 1- (x2 + 1)
∆x
Lim
∆x  0
x2 + 2∆x2 + ∆x2 + 1 – x2 – 1
∆x
Lim
∆x  0
2∆x2 + ∆x2
∆x
Lim
∆x  0
∆x (2x + ∆x)
∆x
Lim
∆x  0
(2x + ∆x)
(2x + (0)) = 2x
A Few Derivative Rules

1. Constant Rule d/dx [c] = 0
*The derivative of variables such as x and y is 1.*
2. Power Rule d/dx [xn] = nxn-1
This means that you move the exponent to the front
and subtract 1.
Ex. y = x3 is 3x2
3. Sine and Cosine
d/dx [sinx] = cosx
d/dx [cosx] = -sinx
*Additional Trig Techniques*
d/dx [tanx] = sec2x
d/dx [secx] = secx tanx
d/dx [cscx] = - cscx cot
d/dx [cotx] = - csc2x
Anything with a c is negative. *
4. Product Rule d/dx [ f(x) g(x)] = f ‘(x) g(x) + f(x) g’(x)
Ex. (3x +1) * (5 + 4x) * f ‘(x) = 3 and g‘(x) = 4*
f(x)
g(x)
3 * (5 + 4x) + (3x +1) * 4
f ‘(x) g(x) + f(x) * g ‘(x)
15 + 12x +12x + 4 = 24x + 19
5. Quotient Rule
d/dx [ f(x)/ g(x)] = g (x) f ‘(x) – f(x) g’(x)
[g(x)]2
* For memorization: lo di hi – hi di lo over lo lo *
Ex. 5x – 2
* f ‘(x) = 5 and g ‘(x) = 2x
2
x +1
[x2 + 1 * 5] – [5x – 2 * 2x] = [5x2 + 5] – [10x2 – 4x]
(x2 + 1)2
(x2 + 1)2
= - 5x2 – 4x + 5
(x2 + 1)2
6. Chain Rule d/dx = f ‘ [g(x)] * g’ (x)
This means that you multiply the original function by the
derivative of the inside function.
Ex. (2x2 + 5)7 (Power rule: move 7 to front and subtract 1.)
7(2x2 + 5)6 * 4x (Derivative of the inside is multiplied,
inside left intact)
2
6
28x(2x + 5)
Implicit Differentiation

When you implicitly differentiate y variables require dy/dx.
Ex. d/dx [y3] = 3y2 dy/dx
Guidelines
1. Differentiate both sides with respect to x
2. Collect all terms involving dy/dx
3. Factor dy/dx to the left side
4. Solve for dy/dx
Ex. y3 + y2 – 5y – x2 = - 4
3y2 dy/dx + 2y dy/dx – 5 dy/dx – 2x = 0
dy/dx (3y2 + 2y – 5) = 2x
dy/dx =
2x
3y2 + 2y – 5
Rate of Change

Rate = distance
Average Velocity = ∆d
time
∆t
S (t) = Position Function
S ‘(t) = Velocity (Find the derivative)
V ‘(t) = Acceleration (Find the second derivative)
A(t) = S ‘’(t)
Ex. S (t) = 3x2 + 3
V(t) = 6x
A(t) = 6
Chapter 3:
Antiderivatives

Antiderivatives vs. Derivatives
Definite Integrals vs. Indefinite

Antiderivative: a function F is an antiderivative of f on an
interval I if F’(x) = f(x) for all x in I.
An antiderivative is truly the inverse operation of the
derivative.
A definite integral expresses the difference between the
upper and lower limits of the function. Example:
An indefinite integral does not include the limits of the
function. Example:
Basic Integration Rules

Accumulation Functions

Accumulation functions shows how you can figure out the
velocity and position of a function when you’re given the
acceleration. You can find the velocity by finding the
antiderivative of the acceleration. Then, to find the position
of the object you have to find the antiderivative of the
velocity.
Ex. a(t) = 6 sec/min2, x(0) = 2 and v(0) = 5
v(t) =  6dt
v(t) = 6t + c (plug in initial condition to solve for c)
s(t) =  6t  5dt
s(t) = 6t + 5t + c (plug in initial condition again)
s(t) = 6t+5t+2
Chapter 4: Application
Problem

How Calculus Relates to
Physics

Calculus can be applied to Physics in terms of velocity,
acceleration and position.
Ex. A model rocket is fired vertically upward from rest. Its
acceleration for the first three seconds is a(t) = 60t at which time
the fuel is exhausted and it becomes a freely “falling” body.
Fourteen seconds later, the rocket’s parachute opens, and the
(downward) velocity slows linearly to -18ft/second in 5
seconds. The rocket then “floats” to the ground at that rate.
At what time does the rocket reach its maximum height?
The rocket accelerates for the first three seconds. Hence, the rocket
reaches its maximum height at t = 3 seconds.
What is the height?
A(t) = 60t therefore to find the velocity you find the antiderivative
of a(t). V(t) = 30t2 +c. To find the height you must find the
antiderivative of the velocity D(t) = 10t2 +C. Using three
seconds as your maximum height, D(3) = 270 meters.