The document defines the tangent line and derivative of a function. It provides examples of finding the slope of the tangent line and derivative for various functions at given points using limits. Specifically, it shows how to calculate the slope of the tangent line to f(x)=2x^3 - 1 at (2,15), finding the equation of the tangent line to f(x)=x^2 at (1,1), and calculating the derivative of f(x)=x^2.

1. Based on the activity exercise above, as the distance of the P(x, y) and (1,1) approaches to zero, the
secant line becomes a tangent line that passes through the point (1,1). If we take the limit of the slope of
this secant line as the distance between P(x, y) and (1,1) approaches zero, we determine the slope of the
tangent line at (1,1)
DEFINITION OF THE TANGENT LINE TO THE GRAPH OF A FUNCTION AT A POINT
The slope of the tangent line to the graph of 𝑓 at the point 𝑃(𝑥, 𝑓(𝑥)) is given by the formula below if
the limit exists.
𝒎 = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉
EXAMPLE #1: Find the equation of the tangent line to the graph of 𝑓(𝑥) = 2𝑥3
− 1 at a point (2,15)
SOLUTION EXPLANATION
𝒎 = 𝒍𝒊𝒎
𝒉→𝟎
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉
Write the formula
𝒍𝒊𝒎
𝒉→𝟎
𝒇(𝟐 + 𝒉) − 𝒇(𝟐)
𝒉
Substitute the 𝑥 with 2 from the given
coordinates (2,15)
𝒍𝒊𝒎
𝒉→𝟎
𝟐(𝟐 + 𝒉)𝟑
− 𝟏 − [𝟏𝟓]
𝒉
Substitute the following values the function
Note: use evaluating a function
1st
value 𝒙 = 𝟐 + 𝒉
𝑓(2 + ℎ) = 2𝑥3
− 1
𝑓(2 + ℎ) = 2(2 + ℎ)3
− 1
Points on 𝒇(𝒙) = 𝒙𝟐 Slope of the Secant Line as
𝑷(𝒙, 𝒚) gets closer to (𝟏, 𝟏)
(2.5, 6.25) & (1,1) 3.5
(2, 4) & (1,1) 3
(1.4, 1.96) & (1,1) 2.4
(1.25, 1.56) & (1,1) 2.24
(1.1, 1.21) & (1,1) 2.21
(1.01, 1.02) & (1,1) 2

2. 𝟐nd
value 𝒙 = 𝟐
𝑓(2) = 2𝑥3
− 1
𝑓(2) = 2(2)2
− 1
𝑓(2) = 2(8) − 1
𝑓(2) = 16 − 1
𝑓(2) = 15
𝒍𝒊𝒎
𝒉→𝟎
[𝟐(𝟐 + 𝒉)(𝟐 + 𝒉)(𝟐 + 𝒉) − 𝟏] − [𝟏𝟓]
𝒉
Expand the (2 + ℎ)3
𝒍𝒊𝒎
𝒉→𝟎
[𝟐(𝒉𝟑
+ 𝟔𝒉𝟐
+ 𝟏𝟐𝒉 + 𝟖) − 𝟏] − [𝟏𝟓]
𝒉
Algebraic manipulation of the polynomial
(multiplication):
(𝟐 + 𝒉)(𝟐 + 𝒉)(𝟐 + 𝒉)
(𝟒 + 𝟒𝒉 + 𝒉𝟐
)(𝟐 + 𝒉)
(𝒉𝟑
+ 𝟔𝒉𝟐
+ 𝟏𝟐𝒉 + 𝟖)
𝒍𝒊𝒎
𝒉→𝟎
𝟐𝒉𝟑
+ 𝟏𝟐𝒉𝟐
+ 𝟐𝟒𝒉 + 𝟏𝟔 − 𝟏 − 𝟏𝟓
𝒉
Multiply the 2 to the polynomial ℎ3
+ 6ℎ2
+
12ℎ + 8
𝒍𝒊𝒎
𝒉→𝟎
𝟐𝒉𝟑
+ 𝟏𝟐𝒉𝟐
+ 𝟐𝟒𝒉
𝒉
Algebraic manipulation: 16 − 1 − 15 = 0
𝒍𝒊𝒎
𝒉→𝟎
𝒉(𝟐𝒉𝟐
+ 𝟏𝟐𝒉 + 𝟐𝟒)
𝒉
𝒍𝒊𝒎
𝒉→𝟎
𝟐𝒉𝟐
+ 𝟏𝟐𝒉 + 𝟐𝟒
Factor the equation and cancel the ℎ
𝟐(𝟎)𝟐
+ 𝟏𝟐(𝟎) + 𝟐𝟒 Substitute the value of ℎ = 0
24 Final Answer
EXAMPLE #2: Find the equation of the tangent line to the graph of 𝑓(𝑥) = 𝑥2
at a point (1,1)
SOLUTION EXPLANATION
𝒎 = 𝒍𝒊𝒎
𝒉→𝟎
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉
Write the formula

3. 𝒍𝒊𝒎
𝒉→𝟎
𝒇(𝟏 + 𝒉) − 𝒇(𝟏)
𝒉
Substitute the 𝑥 with 1 from the given
coordinates (1,1)
𝒍𝒊𝒎
𝒉→𝟎
(𝟏 + 𝒉)𝟐
− 𝟏
𝒉
Substitute the following values the function
Note: use evaluating a function
1st
value 𝒙 = 𝟏 + 𝒉
𝑓(1 + ℎ) = 𝑥2
𝑓(1 + ℎ) = (1 + ℎ)2
𝟐nd
value 𝒙 = 𝟏
𝑓(1) = 𝑥2
𝑓(1) = (1)2
𝑓(1) = 1
𝒍𝒊𝒎
𝒉→𝟎
(𝟏 + 𝒉)(𝟏 + 𝒉) − 𝟏
𝒉
Expand the (𝟏 + 𝒉)𝟐
𝒍𝒊𝒎
𝒉→𝟎
𝟏 + 𝟐𝒉 + 𝒉𝟐
− 𝟏
𝒉
Use the FOIL Method
𝒍𝒊𝒎
𝒉→𝟎
𝒉(𝟐 + 𝒉)
𝒉
𝒍𝒊𝒎
𝒉→𝟎
𝟐 + 𝒉
Factor the equation and cancel the ℎ
𝟐 + 𝟎 Substitute the value of ℎ = 0
2 Final Answer
DEFINITION OF THE DERIVATIVE OF A FUNCTION
The derivative of a function with respect to 𝑥 is the function 𝑓′(𝑥) – read as “𝑓 prime of 𝑥” defined by:
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉

4. NOTE: Every time you get the derivative of a function, you are actually determining the slope of the
tangent line at point 𝑥 = 𝑎. The other notations for the derivative of a function is
𝑑𝑦
𝑑𝑥
.
EXAMPLE #3: Find the derivative of 𝑓(𝑥) = 𝑥2
SOLUTION EXPLANATION
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉
Write the formula
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
(𝒙 + 𝒉)𝟐
− 𝒙𝟐
𝒉
Substitute the following value of the function
Note: use evaluating a function
1st
value 𝒙 = 𝒙 + 𝒉
𝑓(𝑥 + ℎ) = 𝑥2
𝑓(𝑥 + ℎ) = (𝑥 + ℎ)2
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
(𝒙 + 𝒉)(𝒙 + 𝒉) − 𝒙𝟐
𝒉
Expand the (𝑥 + ℎ)2
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
(𝒙 + 𝒉)(𝒙 + 𝒉) − 𝒙𝟐
𝒉
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝒙𝟐
+ 𝟐𝒉 + 𝒉𝟐
− 𝒙𝟐
𝒉
𝒇′(𝒙) = 𝐥𝐢𝐦
𝒉→𝟎
𝟐𝒉 + 𝒉𝟐
𝒉
Use the FOIL Method on the polynomial: (𝑥 +
ℎ)(𝑥 + ℎ) and cancel out 𝑥2
𝒍𝒊𝒎
𝒉→𝟎
𝒉(𝟐 + 𝒉)
𝒉
𝒍𝒊𝒎
𝒉→𝟎
𝟐 + 𝒉
Factor the equation and cancel the ℎ
𝟐 + 𝟎 Substitute the value of ℎ = 0
2 Final Answer
Therefore, the derivative of 𝒇 is 2