José Luis Gómez-Muñoz Tecnológico de Monterrey Campus...

José Luis Gómez-Muñoz

Tecnológico de Monterrey Campus Estado de México

THE DERIVATIVE

José Luis Gómez-Muñoz 2

EXERCISE: Evaluate and sketch� � =�

�� +

�

�� +

��

��

� � �

0 f(0)=

1 f(1)=

2 f(2)=

3 f(3)=

4 f(4)=

5 f(5)=

6 f(6)=

7 f(7)=

8 f(8)=

9 f(9)=

10 f(10)=


Answer: � � =�

�� +

�

�� +

��

��

� � �

0 f(0)=0.94

1 f(1)=2.00

2 f(2)=2.94

3 f(3)=3.75

4 f(4)=4.44

5 f(5)=5.00

6 f(6)=5.44

7 f(7)=5.75

8 f(8)=5.94

9 f(9)=6.00

10 f(10)=5.94


Cross-country skiing on a function


Grade (slope) calculation for the skier


Grade (slope) measurement


Example of grade (slope) calculation 1

� =∆�

∆�=3.5 − 2

8 − 3=1.5

5= 0.30 → %&'() = 30%

� =&+,)

&-.=∆�

∆�



∆�

∆�=3.8 − 1.4

9 − 1=2.4

8= 0.30 → %&'() = 30%

� =&+,)

&-.=∆�

∆�



� =∆1

∆2=

3.� 3.��

��=

4.55

5= −0.11

%&'() = 11% downwards


Lets calculate the grade (slope) for our skier


First slope calculation for � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10

∆�

∆�=� 10 − � 0

10 − 0=5.9375 − 0.9375

10 − 0= 0.5

∆�

∆�=� 10 − � 0

10 − 0

� � =−1

16� +

9

8� +

15

16

� =&+,)

&-.=∆�

∆�


Is this slope missleading?

∆�

∆�=5.9375 − 0.9375

10 − 0= 0.5

∆�

∆�=� 10 − � 0

10 − 0

� � =−1

16� +

9

8� +

15

16

� =&+,)

&-.=∆�

∆�

Is this value useful for the skier?


�� =� 5 − � 0

5 − 0

�� = 0.8125

� =� 10 − � 5

10 − 5

� = 0.1875

Improved slope calculation � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =� 0 + 5 − � 0

5

�� = 0.8125

� =� 5 + 5 − � 5

5

� = 0.1875

∆� = 5; � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =� 0 + 5 − � 0

5= 0.8125

�� = 0.8125 is the slope for the interval 0 ≤ � ≤ 5

� = 2.5 is the middle point of that interval 0 ≤ � ≤ 5

We set a point at 2.5, 0.8125

� =� 5 + 5 − � 5

5= 0.1875

� = 0.1875 is the slope for the interval 5 ≤ � ≤ 10

� = 7.5 is the middle point of interval 5 ≤ � ≤ 10

We set a point at 7.5, 0.1875

Graph the two points →

Slopes in another graph � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =: .� : 4

.�4= 0.97

� =: � : .�

� .�= 0.66

�3 =: 5.� : �

5.��= 0.34

�; =: �4 : 5.�

�45.�= 0.03

Third slope calculation � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =: 4< .� : 4

.�= 0.97

� =: .�< .� : .�

.�= 0.66

�3 =: �< .� : �

.�= 0.34

�; =: 5.�< .� : 5.�

.�= 0.03

∆� = 2.5; � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


: 4< .� : 4

.�= 0.97 → 1.2, 0.97

: .�< .� : .�

.�= 0.66 → 3.7, 0.66

: �< .� : �

.�= 0.34 → 6.2, 0.34

: 5.�< .� : 5.�

.�= 0.03 → 8.7, 0.03

Graph the points →

Slopes in another graph � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =: < :

=

� =: < :

=

�3 =: < :

=

�; =: < :

=

�� =: < :

=

EXERCISE: � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10

2. Graph the slope values →

1. Use ∆� = 2.00, complete and calculate:


�� =: 4< : 4

= 1.00

� =: < :

= 0.75

�3 =: ;< : ;

= 0.50

�; =: �< : �

= 0.25

�� =: �< : �

= 0.00

Answers: � � =�

�� +

�

�� +

��

��; 0 ≤ � ≤ 10


�� =: < :

=

� =: < :

=

�3 =: < :

=

�; =: < :

=

EXERCISE: � � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10




Answers: � � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10

�� =: 4< .� : 4

.�= 0.09

� =: .�< .� : .�

.�= 1.34

�3 =: �< .� : �

.�= 0.76

�; =: 5.�< .� : 5.�

.�= 0.03


� � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10

�� =: 4< .� : 4

.�= 0.09

� =: .�< .� : .�

.�= 1.34

�3 =: �< .� : �

.�= 0.76

�; =: 5.�< .� : 5.�

.�= 0.03

All these calculations follow this pattern:

∆�

∆�=� � + ℎ − � �

ℎ

Here ℎ = 2.5


� � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10 with ℎ = ∆� = 2

Here is the graph for:

∆�

∆�=� � + ℎ − � �

ℎ

with ℎ = 2


� � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10 with ℎ = ∆� = 1.25


∆�

∆�=� � + ℎ − � �

ℎ

with ℎ = 1.25


� � =�4

� ��=>?� @⁄ <�+

�

�; 0 ≤ � ≤ 10 with ℎ = ∆� = 1.00


∆�

∆�=� � + ℎ − � �

ℎ

with ℎ = 1.00


Smaller ℎ gives more precise information


∆�

∆�=� � + ℎ − � �

ℎ

with ℎ = 0.5

We know more detail about the “slope” of the

curve at each point if ℎ is smaller, that is, if ℎ

approaches zero, ℎ → 0


Smaller ℎ → 0 gives exact information about the slope

∆�

∆�=� � + ℎ − � �

ℎ

Exact “slope” (called derivative):

(�

(�= lim

F→4

� � + ℎ − � �

ℎ

Approximated slope:


We will sketch a good approximation to the derivative of this

function, without doing any calculation


Sketch a good approximation to the derivative, step 1:


Sketch a good approximation to the derivative, answer:


EXERCISE: Sketch a good approximation to the derivative of the



Answer: Sketch of the derivative


Expression for the approximated derivative

∆�

∆�=� � + ℎ − � �

ℎ

We will find an expression for the

approximated derivative ∆1

∆2of the

function � � =2@

��, such that it can be

used for any value of ℎ.

Approximated derivative:


Some algebra to obtain the expression:

∆�

∆�=� � + ℎ − � �

ℎ

∆�

∆�=

� + ℎ

16−�

16ℎ

∆�

∆�=

� + ℎ − �

16ℎ1

→

Let � � =2@

��

∆�

∆�=

� + ℎ − �

16ℎ

�� = 0 → 0 + 2 − 0

16 2= 0.125 = ��

� = 2 → 2 + 2 − 2

16 2= 0.375 = �

�3 = 4 → 4 + 2 − 4

16 2= 0.625 = �3

�; = 6 → 6 + 2 − 6

16 2= 0.875 = �;

�� = 8 → 8 + 2 − 8

16 2= 1.125 = ��


Example with ℎ = 2

Let � � =2@

��

∆�

∆�=

� + ℎ − �

16ℎ

Let ℎ = 2

�� = 0 → 0 + 2 − 0

16 2= 0.125 = ��

� = 2 → 2 + 2 − 2

16 2= 0.375 = �

�3 = 4 → 4 + 2 − 4

16 2= 0.625 = �3

�; = 6 → 6 + 2 − 6

16 2= 0.875 = �;

�� = 8 → 8 + 2 − 8

16 2= 1.125 = ��


Example with ℎ = 2 and graph

Let � � =2@

��

∆�

∆�=

� + ℎ − �

16ℎ

Let ℎ = 2

�� = 0 → 0 + 1 − 0

16 1= 0.0625 = ��

� = 1 → 1 + 1 − 1

16 1= 0.1875 = �

�3 = 2 → 2 + 1 − 2

16 1= 0.3125 = �3

�; = 3 → 3 + 1 − 3

16 1= 0.43775 = �;

�� = 4 →4 + 1 − 4

16 1= 0.5625 = ��

�� = 5 → 5 + 1 − 5

16 1= 0.6875 = ��

�5 = 6 → 6 + 1 − 6

16 1= 0.8125 = �5

�� = 7 → 7 + 1 − 7

16 1= 09375 = ��

�� = 8 → 8 + 1 − 8

16 1= 1.0625 = ��

��4 = 9 → 9 + 1 − 9

16 1= 1.875 = ��4


Example with ℎ = 1

Let � � =2@

��

∆�

∆�=

� + ℎ − �

16ℎ

Let ℎ = 1


This expression cannot be used with ℎ = 0

Let � � =2@

��

∆�

∆�=

� + ℎ − �

16ℎ

We know more detail about the “slope”

of the curve at each point if ℎ is smaller,

however, this expression cannot be

used with the samallest possible value

of ℎ, which is ℎ = 0.

For example, with � = 1 and ℎ = 0 →1 + 0 − 1

16 0=0

0= G&&H&!


Enter limits and derivatives:

Let � � =2@

��

J1

J2= lim

F→4

KLM @

NO

K@

NO

F

J1

J2= lim

F→4

2<F @2@

��F

J1

J2= lim

F→4

2@< 2F<F@2@

��F

J1

J2= lim

F→4

2F<F@

��F= lim

F→4

2<F F

��F

J1

J2= lim

F→4

2<F

��=

2

��=

2

�→

J1

J2=

2

�


The derivative can be evaluated at any value of �

y =�

16

J1

J2=

2

�

José Luis Gómez-Muñoz

EXERCISES


�� =: < :

�. �=

� =: < :

�. �=

�3 =: < :

�. �=

�; =: < :

�. �=

�� =: < :

�. �=

�� =: < :

�. �=

�5 =: < :

�. �=

�� =: < :

�. �=

EXERCISE: � � =�

� +

�

;� ; 0 ≤ � ≤ 10




�� =: 4.44<�. � : 4.44

�. �= 2.19

� =: �. �<�. � : �. �

�. �= 1.56

�3 =: .�4<�. � : .�4

�. �= 0.94

�; =: 3.5�<�. � : 3.5�

�. �= 0.31

�� =: �.44<�. � : �.44

�. �= −0.31

�� =: �. �<�. � : �. �

�. �= −0.94

�5 =: 5.�4<�. � : 5.�4

�. �= −1.56

�� =: �.5�<�. � : �.5�

�. �= −2.19

Answers: � � =�

� +

�

;� ; 0 ≤ � ≤ 10

END


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