The document discusses key concepts related to calculus including: - The definition of a derivative as the instantaneous rate of change of a function, obtained by taking the limit of the average rate of change as the change in x approaches 0. - Techniques for finding derivatives including differentiation rules for basic functions. - Relationship between a function's derivative and whether it is increasing or decreasing over an interval. - Concepts of local/global extrema and how to analyze a function's critical points and inflection points. - Using optimization techniques like taking derivatives to find maximum/minimum values of expressions subject to constraints.

3.  Gradient/slope of the line gives the
inclinational of the line.
 Slope of a line joining the points
(x1,y1) and (x2,y2) is given by
slo𝑝𝑒 =
(𝑦2−𝑦1)
(𝑥2−𝑥1)
 In the equation 𝑦 = 𝑚𝑥 + 𝑐
 m is the slope of the line of equation
 c is the y-intercept.
 The sum of an infinite geometric
series with first term 𝑢1 and common
ratio 𝑟 (|𝑟| < 1) is
𝑛=1
∞
𝑢𝑛 =
𝑢1
1 − 𝑟

4.  Limit of a function provides what value the
output of a function approaches as input
approaches a certain value.
lim
𝑥→𝑎
𝑓 𝑥 = 𝐿
 𝑓(𝑥) approaches value 𝐿 as 𝑥 approaches 𝑎
(both from the right and left)
lim
𝑥→𝑎+
𝑓 𝑥 = 𝑃
 𝑓(𝑥) approaches value 𝑃 as 𝑥 approaches 𝑎
from the right
lim
𝑥→𝑎−
𝑓 𝑥 = 𝑄
 𝑓(𝑥) approaches value 𝑄 as 𝑥 approaches 𝑎
from the left
x
y
Graph of 𝑓 𝑥 =
1
𝑥

5.  If lim
𝑥→𝑎+
𝑓 𝑥 = 𝐿 and lim
𝑥→𝑎−
𝑓 𝑥 = 𝐿
then limit exists and we say
lim
𝑥→𝑎
𝑓 𝑥 = 𝐿
lim
𝑥→𝑎−
𝑓 𝑥 = 𝑄
x
y

6. lim
𝑥→+∞
𝑓 𝑥 = 𝑝
 As 𝑥 approaches +∞, 𝑓(𝑥)
approaches 𝑝
lim
𝑥→−∞
𝑓 𝑥 = 𝑞
 As 𝑥 approaches −∞, 𝑓(𝑥)
approaches 𝑞
 Lines y = p and y = q are called
horizontal asymptotes of function
𝑓(𝑥)
Graph of 𝑓 𝑥 =
3𝑥
𝑥−5

7. lim
𝑥→𝑐
𝑓 𝑥 = ±∞
 As 𝑥 approaches 𝑐 from the left or
the right, 𝑓(𝑥) approaches +∞ 𝑜𝑟 −
∞
 Line 𝑥 = 𝑐 is called the vertical
asymptote of function 𝑓(𝑥)
Graph of 𝑓 𝑥 =
3𝑥
𝑥−5

8. 𝑆𝑛 = 1
𝑛
𝑢𝑟 =
𝑢1(1−𝑟𝑛)
1−𝑟
(sum of n temrs of a geometric series)
 If 𝑟 < 1, then lim
𝑛→∞
𝑆𝑛 =
𝑢1
1−𝑟
 The sum of infinite geometric series
converges if 𝑟 < 1, else diverges

10.  Average rate of change of a function
is provided by the gradient/slope of
the function.
∆𝑦
∆𝑥
=
(𝑦2 − 𝑦1)
(𝑥2 − 𝑥1)
=
𝑓(𝑥2) − 𝑓(𝑥1)
(𝑥2 − 𝑥1)
 Assume 𝑥1 = 𝑥 & 𝑥2 = 𝑥 + ℎ
∆𝑦
∆𝑥
=
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
(𝑥 + ℎ − 𝑥)
=
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
 Instantaneous rate of change of a
function is the derivative of the
function.
 This occurs when ℎ is really close to
0.
f′
x =
𝑑𝑦
𝑑𝑥
= lim
ℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
 Differentiation is the method of
obtaining a derivative

11.  Average rate of change of a function
is provided by the gradient/slope of
the function.
∆𝑦
∆𝑥
=
(𝑦2 − 𝑦1)
(𝑥2 − 𝑥1)
=
𝑓(𝑥2) − 𝑓(𝑥1)
(𝑥2 − 𝑥1)
 Assume 𝑥1 = 𝑥 & 𝑥2 = 𝑥 + ℎ
∆𝑦
∆𝑥
=
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
(𝑥 + ℎ − 𝑥)
=
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
 Instantaneous rate of change of a
function is the derivative of the
function.
 This occurs when ℎ is really close to
0.
f′
x =
𝑑𝑦
𝑑𝑥
= lim
ℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
 Differentiation is the method of
obtaining a derivative

12.  ℝ - real numbers, 𝜖 – belongs to
 If 𝑓 𝑥 = 𝑥𝑛, where 𝑛 𝜖 ℝ, then
𝑓′ x = 𝑛𝑥𝑛−1
 If 𝑓 𝑥 = 𝑐, where c 𝜖 ℝ, then 𝑓′ x =
0
 𝑐𝑓 𝑥 ′ = 𝑐𝑓′ 𝑥 , where 𝑐 𝜖 ℝ
 𝑓 𝑥 ± 𝑔 𝑥 ′ = 𝑓′ 𝑥 ± 𝑔′ 𝑥

13.  If the slope of a tangent is 𝑚, then
the slope of the normal is −
1
𝑚

15.  Let 𝑓 𝑥 = 𝑢 and 𝑔 𝑥 = 𝑣
 If 𝑦 = 𝑓𝑜𝑔 𝑥 = 𝑓 𝑔(𝑥) = 𝑓(𝑣), then
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑣
∗
𝑑𝑣
𝑑𝑥
= 𝑓′ 𝑣 ∗ 𝑣′
or
𝑑𝑦
𝑑𝑥
= 𝑓′(𝑔(𝑥)) ∗ 𝑔′(𝑥)

16.  Let 𝑓 𝑥 = 𝑢 and 𝑔 𝑥 = 𝑣
 If 𝑦 = 𝑓 𝑥 ∗ 𝑔 𝑥 = uv, then
𝑑𝑦
𝑑𝑥
= 𝑢𝑣 ′ = 𝑢′𝑣 + 𝑢𝑣′
or
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
∗ 𝑣 + 𝑢 ∗
𝑑𝑣
𝑑𝑥
or
𝑑𝑦
𝑑𝑥
= 𝑓′
𝑥 ∗ 𝑔 𝑥 + 𝑓 𝑥 ∗ 𝑔′(𝑥)

17.  Let 𝑐and 𝑔 𝑥 = 𝑣 (≠ 0)
 If 𝑦 =
𝑓(𝑥)
𝑔(𝑥)
=
𝑢
𝑣
, then
𝑑𝑦
𝑑𝑥
=
𝑢
𝑣
′
=
𝑢′𝑣 − 𝑢𝑣′
𝑣2
or
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
𝑣 − 𝑢
𝑑𝑣
𝑑𝑥
𝑣2
or
𝑑𝑦
𝑑𝑥
=
𝑓′ 𝑥 𝑔 𝑥 − 𝑓 𝑥 𝑔′(𝑥)
[𝑔(𝑥)]2

18. First and Second Derivatives

19.  Increasing Function
 A function is increasing in an interval
if y value increases as x value
increases.
 This means slope/gradient is positive
in the interval.
𝑓′ 𝑥 > 0, 𝑎 < 𝑥 < 𝑏, then function 𝑓 𝑥
is increasing in the interval ]𝑎, 𝑏[
 Decreaseing Function
 A function is decreasing in an
invterval if y value decreases as x
value increases.
 This means slope/gradient is negative
in the interval.
𝑓′(𝑥) < 0, 𝑎 < 𝑥 < 𝑏, then the function
𝑓 𝑥 is decreasing in the interval ]𝑎, 𝑏[

20.  Global Minimum – Point A, which
has the least y value in the domain
 Global Maximum – Point D, which
has the highest y value in the
domain
 Turning Points – x-value 𝑐 is a
turning point if 𝑓′ 𝑐 = 0
 Local Minimum – Point C, which has
the least y value in the domain
 Local Maximum – Point B, which
has the least y value in the domain
B
A
C
D

21.  Local Min – Point C
Point 𝑐 is a local min, if 𝑓′ 𝑐 = 0,
𝑓′ 𝑐 < 0 for 𝑥 < 𝑐 & 𝑓′ 𝑐 > 0 for 𝑥 >
𝑐
 Local Max – Point B
Point 𝑏 is a local min, if 𝑓′ 𝑏 = 0,
𝑓′ 𝑏 > 0 for 𝑥 < 𝑏 & 𝑓′ 𝑏 < 0 for 𝑥 >
𝑏
B
A
C
D

22.  𝑓′ 𝑥 =
𝑑𝑦
𝑑𝑥
, 𝑓′′ 𝑥 =
𝑑2𝑦
𝑑𝑥2
 Local Min – Point C
if 𝑓′ 𝑐 = 0 & 𝑓′′
𝑐 > 0
 Local Max – Point B
if 𝑓′ 𝑏 = 0 & 𝑓′′ 𝑏 < 0
 if 𝑓′ 𝑏 = 0 & 𝑓′′ 𝑏 = 0, we cannot
conclude whether it’s a local max or
local min
B
A
C
D

23.  Inflexion point – where concavity
changes
 If 𝑐 is an inflexion point of 𝑓 𝑥 , then
𝑓′′ 𝑐 = 0
 𝑓(𝑥) is concave down in the interval
𝑎 < 𝑥 < 𝑏 if 𝑓′′ 𝑥 < 0
 𝑓 𝑥 is concave up in the interval
𝑎 < 𝑥 < 𝑏 if 𝑓′′ 𝑥 > 0
 If 𝑓′ 𝑐 = 0 and 𝑓′′ 𝑐 = 0, then 𝑥 = 𝑐
is a Horizontal point of inflexion
Concave Down
Concave Up

25. Optimization and Kinematics

26. Steps to solve optimization question.
1. Assume variables to denote the
terms of the
2. Equation or Constraint. Write an
equation in multi variable using the
constraint given.
3. Expresssion or Optimization – write
an expression to be optimized.
4. Convert the expression into a single
variable by using the equation to
make necessary substitution
5. Optimize the expression by
differentiating and equating to zero.
6. Solve the optimization equation for
x. Substitute in the expression to get
the optimized value.

27.  If velocity and acceleration have the same sign, the body is accelerating.
 If velocity and acceleration are in the opposite sign, the body is decelerating.
Displacement
s(t)
Acceleration
a(t)
Velocity
v(t)
𝑣 𝑡 =
𝑑𝑠
𝑑𝑡
= 𝑠′(𝑡) 𝑎 𝑡 =
𝑑𝑣
𝑑𝑡
= 𝑣′(𝑡)
𝑎 𝑡 =
𝑑2𝑠
𝑑𝑡2 = 𝑠′′(𝑡)