1. Derivatives
Days 32-41

2. Warm-up Monday, January 26 – A-Day
lim 𝑥→0 2 𝑥 6 +6 𝑥 3 4 𝑥 5 +3 𝑥 3 is
1 2 1 2
nonexistent
#2 AP Practice Exam – Non-Calculator
lim 𝑥→0 2 𝑥 6 +6 𝑥 3 4 𝑥 5 +3 𝑥 3 = lim 𝑥→0 𝑥 3 2 𝑥 𝑥 3 4 𝑥 = lim 𝑥→0 2 𝑥 𝑥 = 2∙ ∙ = 6 3 =2 D
Mistakes:
B – did lim as x ⟶ ∞
C – assumed 0/0 = 1
E – assumed 0/0 is nonexistent

3. Day 32
Use the definition of derivative to find the derivative of a given function at a given point.

4. Foerster: Exploration 3 – 1
We are not transitioning to finding derivatives by hand using the formal limit definition of derivative at a pont. We have in the first two units visited this idea a couple of times. In this exploration we will keep the functions as quadratics in order to keep the limits manageable algebraically.

5. Spaceship Problem: A spaceship approaches a far-off planet
Spaceship Problem: A spaceship approaches a far-off planet. At time t minutes after its retrorockets fire, its distance, 𝑓 𝑡 kilometers, from the surface.
The figure shows the graph of f. Confirm by grapher that this is correct.
Previously we learned that the instantaneous rate of change (derivative) at a given point could be estimated by finding the average rate of change (slope) between the given point and another point that is very close.
Find the average rate of change of 𝑓 𝑡 with respect to t from t = 5 to 5.1. What are the units of this rate of change?
Chunking: Problems 1-2
Problems 1 and 2 are here to make sure the students have the function entered correctly into the calculator and remind them how to calculate difference quotients efficiently.

6. 𝑓 5.1 −𝑓 5 5.1−5 =2.1 kilometers per minute
Spaceship Problem: A spaceship approaches a far-off planet. At time t minutes after its retrorockets fire, its distance, 𝑓 𝑡 kilometers, from the surface.
The figure shows the graph of f. Confirm by grapher that this is correct.
Previously we learned that the instantaneous rate of change (derivative) at a given point could be estimated by finding the average rate of change (slope) between the given point and another point that is very close.
Find the average rate of change of 𝑓 𝑡 with respect to t from t = 5 to 5.1. What are the units of this rate of change?
𝑓 5.1 −𝑓 −5 =2.1 kilometers per minute
Chunking: Problems 1-2
Problems 1 and 2 are here to make sure the students have the function entered correctly into the calculator and remind them how to calculate difference quotients efficiently.

7. The average rate of change, 𝑚 𝑡 , of 𝑓 𝑡 from 5 to t is 𝑚 𝑡 = 𝑓 𝑡 −𝑓 5 𝑡−5 . By appropriate substitution, express 𝑚 𝑡 in terms of t.
Find the limit of 𝑚 𝑡 in problem 3 as t approaches 5. This is finding the limit of the average rate of change as the second point 𝑡, 𝑓 𝑡 gets closer and closer to 5, 𝑓 5 .
Chunking: Problems 3-6
Now we are going to evaluate a limit of a difference quotient. Check that the students are doing the limit set up and evaluation correctly.

8. 𝑚 𝑡 = 𝑓 𝑡 −𝑓 5 𝑡−5 = 𝑡 2 −8𝑡+18 − 3 𝑡−5 = 𝑡 2 −8𝑡+15 𝑡−5
The average rate of change, 𝑚 𝑡 , of 𝑓 𝑡 from 5 to t is 𝑚 𝑡 = 𝑓 𝑡 −𝑓 5 𝑡−5 . By appropriate substitution, express 𝑚 𝑡 in terms of t.
𝑚 𝑡 = 𝑓 𝑡 −𝑓 5 𝑡−5 = 𝑡 2 −8𝑡+18 − 3 𝑡−5 = 𝑡 2 −8𝑡+15 𝑡−5
Find the limit of 𝑚 𝑡 in problem 3 as t approaches 5. This is finding the limit of the average rate of change as the second point 𝑡, 𝑓 𝑡 gets closer and closer to 5, 𝑓 5 .
lim 𝑡→5 𝑡 2 −8𝑡+15 𝑡−5 = lim 𝑡→5 𝑡−5 𝑡−3 𝑡−5 =2
Chunking: Problems 3-6
Now we are going to evaluate a limit of a difference quotient. Check that the students are doing the limit set up and evaluation correctly.

9. This number is called the derivative of f at t = 5, written as 𝑓 ′ 5 .
What physical quantity does this number represent? What are the units?
What form does 𝑚 5 take if you substitute 5 for t? What word describes this form? Why is 𝑚 5 undefined?

10. This number is called the derivative of f at t = 5, written as 𝑓 ′ 5 .
What physical quantity does this number represent? What are the units?
𝑓 ′ 5 =2 km per min. This value is the velocity of the spaceship at t = 5.
What form does 𝑚 5 take if you substitute 5 for t? What word describes this form? Why is 𝑚 5 undefined?
𝑚 5 = 𝑓 5 −𝑓 5 5−5 = 0 0 which is undefined because of the division by zero and is called an indeterminate form.

11. On the graph plot a line with slope equal to the derivative you calculated on problem 4, passing through the point on the graph of f where t = 5.
What words can you use to describe the relationship between the line in problem 7 and the graph of the function f?
Chunking: Problems 7-10
These problems review concepts previously visited. Make sure the students still understand the connections: function, derivative, tangent line, rate of change…
#9 – Emphasize a point-slope form. The y-intercept is of no importance to the problem because the tangent line is only an approximation close to the point (5, 3). Therefore the slope-intercept form is not useful.

12. The line is a tangent line to the graph at the point (5, 3).
On the graph plot a line with slope equal to the derivative you calculated on problem 4, passing through the point on the graph of f where t = 5.
What words can you use to describe the relationship between the line in problem 7 and the graph of the function f?
The line is a tangent line to the graph at the point (5, 3).
Chunking: Problems 7-10
These problems review concepts previously visited. Make sure the students still understand the connections: function, derivative, tangent line, rate of change…
#9 – Emphasize a point-slope form. The y-intercept is of no importance to the problem because the tangent line is only an approximation close to the point (5, 3). Therefore the slope-intercept form is not useful.

13. Find an equation of the linear function containing the point 5, 𝑓 5 and having the slope equal 𝑓 ′ 5 .
Plot the graph of f and the line in problem 9 on your grapher. Zoom in repeatedly on the point 5,𝑓 5 . Describe the relationship between the line and the graph of f as you zoom in.

14. Find an equation of the linear function containing the point 5, 𝑓 5 and having the slope equal 𝑓 ′ 5 .
𝑦=2 𝑡−5 +3
Plot the graph of f and the line in problem 9 on your grapher. Zoom in repeatedly on the point 5,𝑓 5 . Describe the relationship between the line and the graph of f as you zoom in.
The line and the curve become the same.

15. The fraction 𝑚 𝑥 = 𝑓 𝑥 −𝑓 𝑐 𝑥−𝑐 is called a difference quotient.
The formal Definition of Derivative (derivate at x = c form)
𝑓 ′ 𝑐 = lim 𝑥→𝑐 𝑓 𝑥 −𝑓 𝑐 𝑥−𝑐
Chunking: Definition of a derivative at a point.
State the formal definition of derivative and explain we will be calculating derivatives at a point algebraically, not using the nDeriv( function of the calculator. Students can use the calculator to check their work.

16. Eucalyptus trees grow better with more water
Eucalyptus trees grow better with more water. Scientists on Hawaii, HI (the Big Island), analyzing where to plant trees, found the volume of wood that grows on a square kilometer, in cubic meters, is approximated by 𝑉 𝑟 =0.2 𝑟 2 −20𝑟+600 where r is rainfall in centimeters per year, and 60≤𝑟≤120. Use the definition of derivative above to calculate 𝑉 ′ Using correct units, explain the meaning of 𝑉 ′ in the context of volume of wood from Eucalyptus trees.
Chunking: Problem 11
A context problem: Make sure they have the units and interpretation correct.

17. 𝑉 ′ 100 =20 cubic meters per centimeter.
Eucalyptus trees grow better with more water. Scientists on Hawaii, HI (the Big Island), analyzing where to plant trees, found the volume of wood that grows on a square kilometer, in cubic meters, is approximated by 𝑉 𝑟 =0.2 𝑟 2 −20𝑟+600 where r is rainfall in centimeters per year, and 60≤𝑟≤120. Use the definition of derivative above to calculate 𝑉 ′ Using correct units, explain the meaning of 𝑉 ′ in the context of volume of wood from Eucalyptus trees.
𝑉 ′ 100 = lim 𝑟→100 𝑉 𝑟 −𝑉 100 𝑟−100 = lim 𝑟→ 𝑟 2 −20𝑟+600 − 600 𝑟−100 = lim 𝑟→ 𝑟 2 −20𝑟 𝑟−100 = lim 𝑟→ 𝑟 𝑟−100 𝑟−100 = lim 𝑟→ 𝑟 =20
𝑉 ′ 100 =20 cubic meters per centimeter.
The volume of wood that grows on a square kilometer is increasing at a rate of 20 cubic meters per centimeter when there is rainfall of 100 centimeters per year.
Chunking: Problem 11
A context problem: Make sure they have the units and interpretation correct.

18. Rules, Theorems, and Examples for Finding Derivatives

19. Find an equation of the tangent line to the graph of 𝑓 𝑥 = 𝑥 2 at x = 5.

20. Find an equation of the tangent line to the graph of 𝑓 𝑥 = 𝑥 2 at x = 5.
𝑓 5 = 5 2 =25 ⇒ 5, 25
𝑓 ′ 5 = lim 𝑥→5 𝑓 𝑥 −𝑓 5 𝑥−5 = lim 𝑥→5 𝑥 2 −25 𝑥−5 = lim 𝑥→5 𝑥+5 𝑥−5 𝑥−5 = lim 𝑥→5 𝑥+5 =5+5 =10=𝑚
𝑦−25=10 𝑥−5

21. The definition of derivative at a point can also be written as
𝑓 ′ 𝑎 = lim ℎ→0 𝑓 𝑎+ℎ −𝑓 𝑎 ℎ

22. Compute 𝑓 ′ 3 , where 𝑓 𝑥 = 𝑥 2 −8𝑥

23. Compute 𝑓 ′ 3 , where 𝑓 𝑥 = 𝑥 2 −8𝑥
𝑓 ′ 3 = lim ℎ→0 𝑓 3+ℎ −𝑓 3 ℎ
𝑓 3+ℎ = 3+ℎ 2 −8 3+ℎ =9+3ℎ+3ℎ+ ℎ 2 −24−8ℎ= ℎ 2 −2ℎ−15
𝑓 3 = 3 2 −8∙3=9−24=−15
𝑓 ′ 3 = lim ℎ→0 ℎ 2 −2ℎ−15 − −15 ℎ = lim ℎ→0 ℎ 2 −2ℎ ℎ = lim ℎ→0 ℎ ℎ−2 ℎ = lim ℎ→0 ℎ−2 =0−2=−2= 𝑓 ′ 3

24. Theorem – Derivative of Linear and Constant Functions
If 𝑓 𝑥 =𝑚𝑥+𝑏 is a linear function, then 𝑓 ′ 𝑎 =𝑚 for all a.
If 𝑓 𝑥 =𝑏 is a constant function, then 𝑓 ′ 𝑎 =0 for all a.

25. Find the derivative of 𝑓 𝑥 =9𝑥−5 at x = 2 and x = 5

26. Find the derivative of 𝑓 𝑥 =9𝑥−5 at x = 2 and x = 5
𝑓 ′ 2 =9
𝑓 ′ 5 =9
𝑓 ′ 𝑥 =9

27. Examples of the Power Rule 𝑑 𝑑𝑥 𝑥 2 =2∙ 𝑥 2−1 =2𝑥
𝑑 𝑑𝑥 𝑥 2 =2∙ 𝑥 2−1 =2𝑥
𝑑 𝑑𝑥 𝑥 5 =5∙ 𝑥 5−1 =5 𝑥 4
𝑑 𝑑𝑥 1 𝑥 3 = 𝑑 𝑑𝑥 𝑥 −3 =−3∙ 𝑥 −3−1 =−3 𝑥 −4 = −3 𝑥 4
𝑑 𝑑𝑥 5 𝑥 4 =5∙4∙ 𝑥 4−1 =20 𝑥 3
𝑑 𝑑𝑥 8 𝑥 = 𝑑 𝑑𝑥 8 𝑥 =8∙ 1 2 ∙ 𝑥 1 2 −1 =4 𝑥 − 1 2 = 4 𝑥 = 4 𝑥
Show students these five examples of the power rule. Give them a few minutes to think-pair-share about what they think might be the power rule.

28. Theorem – The Power Rule For all exponents n, 𝑑 𝑑𝑥 𝑥 𝑛 =𝑛 𝑥 𝑛−1
𝑑 𝑑𝑥 𝑥 𝑛 =𝑛 𝑥 𝑛−1
Be sure to clarify that the base is a function of x and not the exponent. The derivative of 𝑦= 2 𝑥 is not 𝑦 ′ =𝑥 2 𝑥−1 .

29. Theorem – Linearity Rules
Assume that f and g are differentiable.
Sum and Difference Rules: f + g and f – g are differentiable
𝑓+𝑔 ′ = 𝑓 ′ + 𝑔 ′ and 𝑓−𝑔 ′ = 𝑓 ′ − 𝑔 ′
Constant Multiple Rule: For any constant c, cf is differentiable
𝑐𝑓 ′ =𝑐 𝑓 ′

30. Find the points on the graph of 𝑓 𝑥 = 𝑥 3 −12𝑥+4 where the tangent line is horizontal.

31. Find the points on the graph of 𝑓 𝑥 = 𝑥 3 −12𝑥+4 where the tangent line is horizontal.
Tangent line horizontal ⟹ slope = 0 ⟹ 𝑓 ′ 𝑥 =0
𝑓 ′ 𝑥 =3 𝑥 2 −12
3 𝑥 2 −12=0
3 𝑥 2 =12
𝑥 2 =4
𝑥=±2
𝑓 2 = 2 3 −12∙2+4=8−24+4=−12
𝑓 −2 = −2 3 −12 −2 +4=−8+24+4=20
2, −12 and −2, 20

32. Calculate 𝑑𝑔 𝑑𝑡 𝑡=1 where 𝑔 𝑡 = 𝑡 −3 +2 𝑡 − 𝑡 − 4 5

33. Calculate 𝑑𝑔 𝑑𝑡 𝑡=1 where 𝑔 𝑡 = 𝑡 −3 +2 𝑡 − 𝑡 − 4 5
𝑔 𝑡 = 𝑡 −3 +2 𝑡 1/2 − 𝑡 − 4 5
𝑑𝑔 𝑑𝑡 = 𝑔 ′ 𝑡 =−3 𝑡 − 𝑡 − − − 𝑡 − 9 5
𝑑𝑔 𝑑𝑡 =−3 𝑡 −4 + 𝑡 − 𝑡 − 9 5
𝑑𝑔 𝑑𝑡 = −3 𝑡 𝑡 1/ 𝑡 9/5
𝑑𝑔 𝑑𝑡 𝑡=1 = 𝑔 ′ 1 = − / /5 =− =− 6 5

34. There is a unique positive real number e with the property
Theorem – The Number e
There is a unique positive real number e with the property
𝑑 𝑑𝑥 𝑒 𝑥 = 𝑒 𝑥
Show the Insane Mathematician power point in the folder called AB Deriv Rules before showing this slide.

35. Find the tangent line to the graph of 𝑓 𝑥 =3 𝑒 𝑥 −5 𝑥 2 at x = 2 and use it to estimate 𝑓 2.2 .

36. Find the tangent line to the graph of 𝑓 𝑥 =3 𝑒 𝑥 −5 𝑥 2 at x = 2 and use it to estimate 𝑓 2.2 .
We first need the point 2,𝑓 2
𝑓 2 =3 𝑒 2 − =3 𝑒 2 −20⇒ 2,3 𝑒 2 −20
We now need to find the slope at x = 2 ⟹ 𝑓 ′ 2 =𝑚
𝑓 ′ 𝑥 =3 𝑒 𝑥 −10𝑥
𝑓 ′ 2 =3 𝑒 2 −10 2 =3 𝑒 2 −20=𝑚≈2.167
Now write the equation using point-slope form
𝑦− 3 𝑒 2 −20 = 3 𝑒 2 −20 𝑥−2
𝑦−2.167=2.167 𝑥−2

37. Theorem – Derivative of Sine and Cosine
The functions 𝑦= sin 𝑥 and 𝑦= cos 𝑥 are differentiable
𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 and 𝑑 𝑑𝑥 cos 𝑥 =− sin 𝑥

38. Theorem – Derivative of the Natural Logarithm
𝑑 𝑑𝑥 ln 𝑓 𝑥 = 𝑓 ′ 𝑥 𝑓 𝑥

39. Find the derivative 𝑦= ln 𝑥 3 +1 𝑦= ln sin 𝑥 𝑦=2 ln 4 𝑥 3 + 𝑥
At this point, the speed quizzes can be given. Give through this unit to keep them fresh with taking derivatives quickly.

40. Find the derivative 𝑦= ln 𝑥 3 +1 𝑦 ′ = 3 𝑥 2 𝑥 3 +1 𝑦= ln sin 𝑥
𝑦 ′ = 3 𝑥 2 𝑥 3 +1
𝑦= ln sin 𝑥
𝑦 ′ = cos 𝑥 sin 𝑥 = cot 𝑥
𝑦=2 ln 4 𝑥 3 + 𝑥
𝑦 ′ =2∙ 12 𝑥 𝑥 − 𝑥 3 + 𝑥 = 24 𝑥 2 + 𝑥 − 𝑥 3 + 𝑥 1 2
At this point, the speed quizzes can be given. Give through this unit to keep them fresh with taking derivatives quickly.

41. Homework:
Page 139 #s 7-12, 17, 18, even, all
Page 167 #s 1, 2
Page 187 #s 4, 5

42. Warm-up Tuesday, January 27 (Snow day); Wednesday, January 28 – B-Day
The function f is continuous on the closed interval [0, 6] and has the values in the table above. The trapezoidal approximation for 0 6 𝑓 𝑥 𝑑𝑥 found with 3 subintervals of equal length is 52. What is the value of k? 2 6 7 10 14 x 2 4 6
f(x) k 8 12
#8 AP Practice Exam – Non-Calculator
0 6 𝑓 𝑥 𝑑𝑥= 𝑘 𝑘 =52
4+𝑘+𝑘+8+20=52
2𝑘+32=52
2𝑘=20
𝑘=10 D
Mistakes:
A. 2 –
B. 6 – Used right Riemann sum
C. 7 –
E. 14 – Used left Riemann sum

43. Go over homework

44. Speed Quiz 1 AB

45. Warm-up Thursday, January 29 – C-Day
𝑓 𝑥 = 𝑥 2 −3𝑥+9 for 𝑥≤2 𝑘𝑥 for 𝑥>2
The function f is defined above. For what value of k, if any, is f continuous at x = 2? 1 2 3 7
No value of k will make f continuous at x = 2.
#3 AP Practice Exam – Non-Calculator
lim 𝑥→2− 𝑓 𝑥 = lim 𝑥→2− 𝑥 2 −3𝑥+9 = 2 2 −3∙2+9=4−6+9=7
lim 𝑥→2+ 𝑓 𝑥 = lim 𝑥→2+ 𝑘𝑥+1 =𝑘 2 +1
2𝑘+1=7
2𝑘=6
𝑘=3
C. 3
Mistakes:
D. 7 – Only plugged 2 in for top function

46. Speed Quiz

47. Number every other line on your paper 1-18 You will have 15 seconds per problem Find the derivative

48. 𝑑 𝑑𝑥 𝑥 3 1

49. 𝑑 𝑑𝑥 3 𝑥 2

50. 𝑑 𝑑𝑥 1 𝑥 2 3

51. 𝑑 𝑑𝑥 𝑥 4/5 4

52. 𝑑 𝑑𝑥 𝑥 −7 5

53. 𝑑 𝑑𝑥 𝑥 6

54. 𝑑 𝑑𝑥 𝑥 8 7

55. 𝑑 𝑑𝑥 𝑥 8

56. 𝑑 𝑑𝑥 4 −2 9

57. 𝑑 𝑑𝑥 𝑥 0.7 10

58. 𝑑 𝑑𝑥 1 𝑥 9 11

59. 𝑑 𝑑𝑥 6 𝑥 12

60. 𝑑 𝑑𝑥 𝜋 13

61. 𝑑 𝑑𝑥 5 𝑒 𝑥 14

62. 𝑑 𝑑𝑥 𝑥 2 −2𝑥 15

63. 𝑑 𝑑𝑥 ln 𝑥 2 16

64. 𝑑 𝑑𝑥 −4 sin 𝑥 17

65. 𝑑 𝑑𝑥 −2 cos 𝑥 18

66. Switch papers with a neighbor and grab a marker

67. 𝑑 𝑑𝑥 𝑥 3 =3 𝑥 2 1

68. 𝑑 𝑑𝑥 3 𝑥 = 1 3 𝑥 −2/3 2

69. 𝑑 𝑑𝑥 1 𝑥 2 =−2 𝑥 −3 3

70. 𝑑 𝑑𝑥 𝑥 4/5 = 4 5 𝑥 −1/5 4

71. 𝑑 𝑑𝑥 𝑥 −7 =−7 𝑥 −8 5

72. 𝑑 𝑑𝑥 𝑥 =− 1 2 𝑥 −3/2 6

73. 𝑑 𝑑𝑥 𝑥 8 =8 𝑥 7 7

74. 𝑑 𝑑𝑥 𝑥=1 8

75. 𝑑 𝑑𝑥 4 −2 =0 9

76. 𝑑 𝑑𝑥 𝑥 0.7 =0.7 𝑥 −0.3 10

77. 𝑑 𝑑𝑥 1 𝑥 9 =−9 𝑥 −10 11

78. 𝑑 𝑑𝑥 6 𝑥 = 1 6 𝑥 −5/6 12

79. 𝑑 𝑑𝑥 𝜋=0 13

80. 𝑑 𝑑𝑥 5 𝑒 𝑥 =5 𝑒 𝑥 14

81. 𝑑 𝑑𝑥 𝑥 2 −2𝑥 =2𝑥−2 15

82. 𝑑 𝑑𝑥 ln 𝑥 2 = 2𝑥 𝑥 2 = 2 𝑥 16

83. 𝑑 𝑑𝑥 −4 sin 𝑥 =−4 cos 𝑥 17

84. 𝑑 𝑑𝑥 −2 cos 𝑥 =2 sin 𝑥 18

85. Product Rule

86. Examples of the Product Rule 𝑓 𝑥 = 𝑥+5 𝑒 𝑥 Incorrect: 𝑓 ′ 𝑥 = 1 𝑒 𝑥
𝑓 𝑥 = 𝑥+5 𝑒 𝑥
Incorrect: 𝑓 ′ 𝑥 = 1 𝑒 𝑥
Correct: 𝑓 ′ 𝑥 = 𝑥+5 𝑒 𝑥 + 𝑒 𝑥 1
𝑔 𝑥 = 3 𝑥 2 +15𝑥 𝑥 3 +49
Incorrect: 𝑔 ′ 𝑥 = 6𝑥 𝑥 2
Correct: 𝑔 ′ 𝑥 = 3 𝑥 2 +15𝑥 𝑥 𝑥 𝑥+15
𝑄 𝑟 = 1−2𝑟 3𝑟+5
Incorrect: 𝑄 ′ 𝑟 = −2 3
Correct: 𝑄 ′ 𝑟 = 1−2𝑟 𝑟+5 −2
Show students these three examples of the product rule. Give them a few minutes to think-pair-share about what they think might be the product rule.

87. Theorem – Product Rule
If f and g are differentiable functions, then fg is differentiable and
𝑓𝑔 ′ 𝑥 =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓 ′ 𝑥
It may be helpful to remember the Product Rule in words: The derivative of a product is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function:
First∙ Second ′ +Second∙ First ′

88. Find the derivative using the Product Rule
ℎ 𝑥 = 𝑥 2 9𝑥+2
𝑦= 2+ 𝑥 −1 𝑥 3/2 +1
𝑓 𝑡 = 𝑡 2 𝑒 𝑡
𝑔 𝜃 =2 𝜃 3 cos 𝜃

89. Find the derivative using the Product Rule
ℎ 𝑥 = 𝑥 2 9𝑥+2
ℎ ′ 𝑥 = 𝑥 𝑥+2 2𝑥 =9 𝑥 𝑥 2 +4𝑥=27 𝑥 2 +4𝑥
𝑦= 2+ 𝑥 −1 𝑥 3/2 +1
𝑦 ′ = 2+ 𝑥 − 𝑥 1/2 + 𝑥 3/2 +1 −1 𝑥 −2 =3 𝑥 1/ 𝑥 −1/2 − 𝑥 − 1 2 − 𝑥 −2 =3 𝑥 𝑥 −1/2 − 𝑥 −2
𝑓 𝑡 = 𝑡 2 𝑒 𝑡
𝑓 ′ 𝑡 = 𝑡 2 𝑒 𝑡 + 𝑒 𝑡 2𝑡 = 𝑡 2 𝑒 𝑡 +2𝑡 𝑒 𝑡 = 𝑒 𝑡 𝑡 2 +2𝑡
𝑔 𝜃 =2 𝜃 3 cos 𝜃
𝑔 ′ 𝜃 = 2 𝜃 3 − sin 𝜃 + cos 𝜃 6 𝜃 2 =−2 𝜃 3 sin 𝜃 +6 𝜃 2 cos 𝜃

90. Examples of the Quotient Rule
Examples of the Quotient Rule. Try to determine the Quotient Rule given the below incorrect and correct worked out answers.
𝑓 𝑥 = 2 𝑥 𝑥+15
Incorrect: 𝑓 ′ 𝑥 = 6 𝑥 2 2
Correct: 𝑓 ′ 𝑥 = 2𝑥 𝑥 2 − 2 𝑥 𝑥 = 12 𝑥 𝑥 2 −4 𝑥 3 −6 2𝑥 = 8 𝑥 𝑥 2 −6 2𝑥+15 2
𝑔 𝑥 = sin 𝑥 3𝑥
Incorrect: 𝑔 ′ 𝑥 = cos 𝑥 3
Correct: 𝑔 ′ 𝑥 = 3𝑥 cos 𝑥 − sin 𝑥 𝑥 2 = 3𝑥 cos 𝑥 −3 sin 𝑥 9 𝑥 2 = 3 𝑥 cos 𝑥 − sin 𝑥 9 𝑥 2 = 𝑥 cos 𝑥 − sin 𝑥 3 𝑥 2
Show students these two examples of the quotient rule. Give them a few minutes to think-pair-share about what they think might be the quotient rule.

91. Theorem – Quotient Rule
If f and g are differentiable functions, then f/g is differentiable for all x such that 𝑔 𝑥 ≠0, and
𝑓 𝑔 ′ 𝑥 = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2
The numerator in the Quotient Rule is equal to the bottom times the derivative of the top minus the top time the derivative of the bottom
Bottom∙ Top ′ −Top∙ Bottom ′ Bottom 2
OR “lo d hi minus hi d lo all over lo lo”
lo d hi−hi d lo lo lo

92. Find the derivative using the Quotient Rule
𝑓 𝑥 = 𝑥 1+ 𝑥 2
𝑔 𝑡 = 𝑒 𝑡 𝑒 𝑡 +𝑡
𝑦= 3 𝑥 2 +𝑥−2 4 𝑥 3 +1

93. Find the derivative using the Quotient Rule
𝑓 𝑥 = 𝑥 1+ 𝑥 2
𝑓 ′ 𝑥 = 1+ 𝑥 − 𝑥 2𝑥 𝑥 = 1+ 𝑥 2 −2 𝑥 𝑥 = 1− 𝑥 𝑥 2 2
𝑔 𝑡 = 𝑒 𝑡 𝑒 𝑡 +𝑡
𝑔 ′ 𝑡 = 𝑒 𝑡 +𝑡 𝑒 𝑡 − 𝑒 𝑡 𝑒 𝑡 𝑒 𝑡 +𝑡 2 = 𝑒 2𝑡 +𝑡 𝑒 𝑡 − 𝑒 2𝑡 − 𝑒 𝑡 𝑒 𝑡 +𝑡 2 = 𝑡 𝑒 𝑡 − 𝑒 𝑡 𝑒 𝑡 +𝑡 2
𝑦= 3 𝑥 2 +𝑥−2 4 𝑥 3 +1
𝑦 ′ = 4 𝑥 𝑥+1 − 3 𝑥 2 +𝑥−2 12 𝑥 𝑥 = 24 𝑥 4 +4 𝑥 3 +6𝑥+1−36 𝑥 4 −12 𝑥 3 −24 𝑥 𝑥 = −12 𝑥 4 −8 𝑥 3 −24 𝑥 2 +6𝑥 𝑥

94. Homework:
Pages #s 1-4, 7, 8, 12, 39-42

95. Warm-up Friday, January 30 – A-Day
The function f is given by 𝑓 𝑥 = 𝑎 𝑥 𝑥 2 +𝑏 . The figure above shows a portion of the graph of f. Which of the following could be the values of the constants a and b?
a = -3, b = 2
a = 2, b = -3
a = 2, b = -2
a = 3, b = -4
a = 3, b = 4
#10 AP Practice Exam – Non-Calculator
Horizontal Asymptote at y = 3 ⇒ lim 𝑥→±∞ 𝑓 𝑥 =3
lim 𝑥→±∞ 𝑎 𝑥 𝑥 = 𝑎 1 =3 ∴𝑎=3
Vertical Asymptote at x = 2 ⟹ lim 𝑥→2 𝑓 𝑥 =±∞
𝑥 2 +𝑏=0 when x = 2
2 2 +𝑏=0 ∴ 𝑏=−4
D. a = 3, b = -4

96. Go Over Homework

97. Derivatives of Trigonometric Functions

98. Exploration
Rewrite the trigonometric function as a quotient then use the quotient rule and the derivatives of sin 𝑥 and cos 𝑥 to find the derivatives of the following functions.
𝑓 𝑥 = tan 𝑥
𝑔 𝑥 = csc 𝑥
ℎ 𝑥 = sec 𝑥
𝑦= cot 𝑥
Break the class into four groups and have each group do one on poster paper.

99. Theorem – Derivatives of Standard Trigonometric Functions
𝑑 𝑑𝑥 tan 𝑥 = sec 2 𝑥
𝑑 𝑑𝑥 sec 𝑥 = sec 𝑥 tan 𝑥
𝑑 𝑑𝑥 csc 𝑥 =− csc 𝑥 cot 𝑥
𝑑 𝑑𝑥 cot 𝑥 =− csc 2 𝑥

100. Find the derivative.
𝑓 𝜃 = tan 𝜃 sec 𝜃
𝑔 𝑥 = 𝑥 2 cot 𝑥
𝑦= csc 𝑥 3𝑒 𝑥

101. Find the derivative.
𝑓 𝜃 = tan 𝜃 sec 𝜃
𝑓 ′ 𝜃 = tan 𝜃 sec 𝜃 tan 𝜃 + sec 𝜃 sec 2 𝜃
= sec 𝜃 tan 2 𝜃 + sec 3 𝜃 = sec 𝜃 tan 2 𝜃 + sec 2 𝜃
𝑔 𝑥 = 𝑥 2 cot 𝑥
𝑔 ′ 𝑥 = 𝑥 2 − csc 2 𝑥 + cot 𝑥 2𝑥
=− 𝑥 2 csc 2 𝑥 +2𝑥 cot 𝑥 =−𝑥 𝑥 csc 2 𝑥 +2 cot 𝑥
𝑦= csc 𝑥 3𝑒 𝑥
𝑦 ′ = 3 𝑒 𝑥 − csc 𝑥 cot 𝑥 − csc 𝑥 3 𝑒 𝑥 𝑒 𝑥 2 = −3 𝑒 𝑥 csc 𝑥 cot 𝑥 − csc 𝑥 9 𝑒 2𝑥 = − csc 𝑥 cot 𝑥 −1 𝑒 𝑥

102. Higher Derivatives

103. Higher derivatives are obtained by repeatedly differentiating a function.
If 𝑓 ′ is differentiable, then the second derivative, denoted 𝑓 ′′ is the derivative
The second derivative is the rate of change of the first derivative.
𝑓 ′′ 𝑥 = 𝑑 𝑑𝑥 𝑓 ′ 𝑥

104. The process of differentiation can be continued, provided that the derivatives exist.
The third derivative, denoted 𝑓 ′′′ 𝑥 or 𝑓 3 𝑥 , is the derivative of 𝑓 ′′ 𝑥 .
In Leibniz notation, we write: 𝑑𝑦 𝑑𝑥 , 𝑑 2 𝑦 𝑑 𝑥 2 , 𝑑 3 𝑦 𝑑 𝑥 3 , 𝑑 4 𝑦 𝑑 𝑥 4 , …
𝑑𝑦 𝑑𝑥 has for its units: unit of y per unit of x
𝑑 2 𝑦 𝑑 𝑥 2 has for its units: units of dy/dx per unit of x, or unit of y per unit of x per unit of x

105. Examples
Calculate 𝑓 ′′′ −1 for 𝑓 𝑥 =3 𝑥 5 −2 𝑥 2 +7 𝑥 −2
Calculate 𝑓 ′′ 𝜋 3 for 𝑓 ′′ 𝑥 =𝑥 cos 𝑥

106. Examples
Calculate 𝑓 ′′′ −1 for 𝑓 𝑥 =3 𝑥 5 −2 𝑥 2 +7 𝑥 −2
𝑓 ′ 𝑥 =15 𝑥 4 −4𝑥−14 𝑥 −3
𝑓 ′′ 𝑥 =60 𝑥 3 −4+42 𝑥 −4
𝑓 ′′′ 𝑥 =180 𝑥 2 −168 𝑥 −5
𝑓 ′′′ 1 =180∙ 1 2 −168∙ 1 −5 =180−168=12
Calculate 𝑓 ′′ 𝜋 3 for 𝑓 ′′ 𝑥 =𝑥 cos 𝑥
𝑓 ′ 𝑥 =𝑥 − sin 𝑥 + cos 𝑥 1 =−𝑥 sin 𝑥 + cos 𝑥
𝑓 ′′ 𝑥 =−𝑥 cos 𝑥 + sin 𝑥 −1 − sin 𝑥 =−𝑥 cos 𝑥 −2 sin 𝑥
𝑓 ′′ 𝜋 3 =− 𝜋 3 cos 𝜋 3 −2 sin 𝜋 3 =− 𝜋 3 ∙ 1 2 −2∙ =− 𝜋 6 − 3

107. Homework:
Page 163 #s 8-24 even, 39, 40
Page 167 #s 6-18 even, 22
Page 168 #s 26, 30, 41, 42, 43

108. Warm-up Monday, February 9 – A-Dayx
𝒇 𝒙
𝒇 ′ 𝒙
𝒈 𝒙
𝒈 ′ 𝒙 1 3 -2 -3 4
The table above gives values of the differentiable functions f and g and their derivatives at x = 1. If ℎ 𝑥 = 2𝑓 𝑥 𝑔 𝑥 , then ℎ ′ 1 =
-28
-16 40 44 47
#89 AP Practice Exam – Calculator
ℎ ′ 𝑥 = 2𝑓 𝑥 +3 𝑔 ′ 𝑥 + 1+𝑔 𝑥 2 𝑓 ′ 𝑥
ℎ ′ 1 = 2𝑓 𝑔 ′ 𝑔 𝑓 ′ 1
ℎ ′ 1 = 2∙ −3 2∙ −2
ℎ ′ 1 = −2 −4 =36+8=44
D. 44

109. Go over work from last week with the sub Constant, Sum, Power, Product, and Quotient Rules Quiz

110. Warm-up Tuesday, February 10 (Snow Day); Wednesday, February 11 (Class Testing); Thurs 2/12 – C-Day
For each function, use the table to evaluate ℎ ′ 2 .
ℎ 𝑥 =𝑓 𝑥 −𝑔 𝑥
ℎ 𝑥 =𝑥𝑓 𝑥
ℎ 𝑥 = 𝑥𝑔 𝑥 3𝑓 𝑥 x
𝒇 𝒙
𝒇 ′ 𝒙
𝒈 𝒙
𝒈 ′ 𝒙 2 3 -3 -2
ℎ ′ 𝑥 = 𝑓 ′ 𝑥 − 𝑔 ′ 𝑥 ; ℎ ′ 2 = 𝑓 ′ 2 − 𝑔 ′ 2 =3− −2 =5
ℎ ′ 𝑥 = 𝑥 𝑓 ′ 𝑥 + 𝑓 𝑥 1 ; ℎ ′ 2 = 2 𝑓 ′ 𝑓 =2∙3+2∙1=6+2=8
ℎ ′ 𝑥 = 3𝑓 𝑥 𝑥 𝑔 ′ 𝑥 + 𝑔 𝑥 1 − 𝑥𝑔 𝑥 3 𝑓 ′ 𝑥 3𝑓 𝑥 2 ; ℎ ′ 2 = 3𝑓 𝑔 ′ 2 +𝑔 2 − 2𝑔 𝑓 ′ 𝑓 = 3∙2 2∙ −2 + −3 −2∙ −3 ∙3∙3 3∙2 2 = 6∙ − = − = = 1 3

111. Go over work from last week with the sub Constant, Sum, Power, Product, and Quotient Rules Quiz
Many of the students did not get all of the work done left, so they worked on it yesterday in class.

112. Chain Rule

113. The Chain Rule is used to differentiate composite functions such as 𝑦= cos 𝑥 3 and 𝑦= 𝑥 4 +1 .

114. Theorem – Chain Rule
If f and g are differentiable, then the composite funcdtion 𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥 is differentiable and
𝑓 𝑔 𝑥 ′ = 𝑓 ′ 𝑔 𝑥 𝑔 ′ 𝑥

115. Example
Calculate the derivative of 𝑦= cos 𝑥 3

116. Example
Calculate the derivative of 𝑦= cos 𝑥 3
𝑦 ′ =− sin 𝑥 3 ∙3 𝑥 2
𝑦 ′ =−3 𝑥 2 sin 𝑥 3

117. Example
Calculate the derivative of 𝑦= 𝑥 4 +1

118. Example
Calculate the derivative of 𝑦= 𝑥 4 +1
𝑦= 𝑥 /2
𝑦 ′ = 𝑥 −1/2 ∙4 𝑥 3
𝑦 ′ =2 𝑥 3 𝑥 −1/2
𝑦 ′ = 2 𝑥 𝑥 4 +1

119. Theorem – General Power and Exponential Rules
If g(x) is differentiable, then
𝑑 𝑑𝑥 𝑔 𝑥 𝑛 =𝑛 𝑔 𝑥 𝑛−1 𝑔 ; 𝑥 for any number n
𝑑 𝑑𝑥 𝑒 𝑔 𝑥 = 𝑔 ′ 𝑥 𝑒 𝑔 𝑥

120. Example
Find the derivative of 𝑦= 𝑥 2 +7𝑥+2 −1/3

121. Example
Find the derivative of 𝑦= 𝑥 2 +7𝑥+2 −1/3
𝑦 ′ =− 𝑥 2 +7𝑥+2 −4/3 2𝑥+7

122. Example
Find the derivative of 𝑦= 𝑒 cos 𝑡

123. Example
Find the derivative of 𝑦= 𝑒 cos 𝑡
𝑦 ′ =− sin 𝑡 𝑒 cos 𝑡

124. Try these on your own. Find the derivative
𝑦= sin 2𝑥+𝜋
𝑓 𝑥 = 9𝑥−2 5
𝑔 𝑥 = 𝑒 7−5𝑥
𝑦= sin 2 𝑥 [Hint: sin2x = (sin x)2]

125. Try these on your own. Find the derivative
𝑦= sin 2𝑥+𝜋
𝑦 ′ = cos 2𝑥+𝜋 ∙2=2 cos 2𝑥+𝜋
𝑓 𝑥 = 9𝑥−2 5
𝑓 ′ 𝑥 =5 9𝑥−2 4 ∙9=45 9𝑥−2 4
𝑔 𝑥 = 𝑒 7−5𝑥
𝑔 ′ 𝑥 = 𝑒 7−5𝑥 ∙−5=−5 𝑒 7−5𝑥
𝑦= sin 2 𝑥 [Hint: sin2x = (sin x)2]
𝑦 ′ =2 sin 𝑥 cos 𝑥

126. Homework
Text page 175 #s 29-33, 43-49
Challenge problem: Find the derivative 𝑦= 𝑥 2 +1

127. Warm-up Tuesday, February 24 – C-Day
If 𝑓 𝑥 = cos 3 4𝑥 , then 𝑓 ′ 𝑥 =
3 cos 2 4𝑥
−12 cos 2 4𝑥 sin 4𝑥
−3 cos 2 4𝑥 sin 4𝑥
12 cos 2 4𝑥 sin 4𝑥
−4 sin 3 4𝑥
#4 AP Practice Exam – Non-Calculator
𝑓 𝑥 = cos 4𝑥 3
𝑓 ′ 𝑥 =3 cos 4𝑥 − sin 4𝑥 4 =−12 cos 2 4𝑥 sin 4𝑥 B

128. Implicit Differentation

129. We have developed the basic techniques for calculating a derivative 𝑑𝑦 𝑑𝑥 when y is given explicitly in terms of x – such as 𝑦= 𝑥 3 +1.
Suppose y is defined implicitly such as 𝑦 4 +𝑥𝑦= 𝑥 3 −𝑥+2 in which the equation cannot be solved for y.
How can we find the slope of the tangent line at a point on the graph?
Although it may be difficult or even impossible to solve for y explicitly as a function of x, we can find 𝑑𝑦 𝑑𝑥 using the method of implicit differentiation.

130. Guidelines for Implicit Differentiation
Differentiate both sides of the equation with respect to x.
The Chain Rule allows us to take the derivative of y by putting 𝑑𝑦 𝑑𝑥 with it.
Collect all terms involving 𝑑𝑦 𝑑𝑥 on the left side of the equation and move all other terms to the right side of the equation.
Factor 𝑑𝑦 𝑑𝑥 out of the left side of the equation.
Solve for 𝑑𝑦 𝑑𝑥 .

131. Example
Find 𝑑𝑦 𝑑𝑥 given that 𝑦 3 + 𝑦 2 −5𝑦− 𝑥 2 =−4

132. Example
Find 𝑑𝑦 𝑑𝑥 given that 𝑦 3 + 𝑦 2 −5𝑦− 𝑥 2 =−4
3 𝑦 2 𝑑𝑦 𝑑𝑥 +2𝑦 𝑑𝑦 𝑑𝑥 −5 𝑑𝑦 𝑑𝑥 −2𝑥=0
3 𝑦 2 𝑑𝑦 𝑑𝑥 +2𝑦 𝑑𝑦 𝑑𝑥 −5 𝑑𝑦 𝑑𝑥 =2𝑥
𝑑𝑦 𝑑𝑥 3 𝑦 2 +2𝑦−5 =2𝑥
𝑑𝑦 𝑑𝑥 = 2𝑥 3 𝑦 2 +2𝑦−5

133. Example
Find 𝑑𝑦 𝑑𝑥 given that 𝑥 3 −𝑥𝑦+ 𝑦 2 =4

134. Example
Find 𝑑𝑦 𝑑𝑥 given that 𝑥 3 −𝑥𝑦+ 𝑦 2 =4
3 𝑥 2 + −𝑥 1 𝑑𝑦 𝑑𝑥 + 𝑦 −1 +2𝑦 𝑑𝑦 𝑑𝑥 =0
3 𝑥 2 −𝑥 𝑑𝑦 𝑑𝑥 −𝑦+2𝑦 𝑑𝑦 𝑑𝑥 =0
−𝑥 𝑑𝑦 𝑑𝑥 +2𝑦 𝑑𝑦 𝑑𝑥 =− 3𝑥 2 +𝑦
𝑑𝑦 𝑑𝑥 −𝑥+2𝑦 =−3 𝑥 2 +𝑦
𝑑𝑦 𝑑𝑥 = −3 𝑥 2 +𝑦 −𝑥+2𝑦 = 3 𝑥 2 −𝑦 𝑥−2𝑦 = 𝑦−3 𝑥 2 2𝑦−𝑥

135. Example
Find an equation of the tangent line at the point P = (1, 1) on the curve 𝑦 4 +𝑥𝑦= 𝑥 3 −𝑥+2

136. Example
Find an equation of the tangent line at the point P = (1, 1) on the curve 𝑦 4 +𝑥𝑦= 𝑥 3 −𝑥+2
4 𝑦 3 𝑑𝑦 𝑑𝑥 +𝑥 𝑑𝑦 𝑑𝑥 +𝑦=3 𝑥 2 −1
𝑑𝑦 𝑑𝑥 = 3 𝑥 2 −1−𝑦 4 𝑦 3 +𝑥
𝑚= −1− = 1 5
𝑦−1= 1 5 𝑥−1

137. Example
Find the slope of the tangent line at the point P = (1, 1) on the graph of 𝑒 𝑥−𝑦 =2 𝑥 2 − 𝑦 2

138. Example
Find the slope of the tangent line at the point P = (1, 1) on the graph of 𝑒 𝑥−𝑦 =2 𝑥 2 − 𝑦 2
𝑒 𝑥−𝑦 1− 𝑑𝑦 𝑑𝑥 =4𝑥−2𝑦 𝑑𝑦 𝑑𝑥
Plug the point in and solve for 𝑑𝑦 𝑑𝑥 would be easier in this case
𝑒 1−1 1− 𝑑𝑦 𝑑𝑥 =4 1 −2 1 𝑑𝑦 𝑑𝑥
1− 𝑑𝑦 𝑑𝑥 =4−2 𝑑𝑦 𝑑𝑥
𝑑𝑦 𝑑𝑥 =3=𝑚

139. Homework
Text page 192 #s 9-12, 15, 18, 21, 23, 25, 31