1. Chapter 3 The Derivative By: Kristen Whaley

2. 3.1 Slopes and Rates of Change  Average Velocity  Instantaneous Velocity  Average Rate of Change  Instantaneous Rate of Change

3. Average Velocity For an object moving along an s-axis, with s= f(t), the average velocity of an object between times t 0 and t 1 is: V average = f(t1) – f(t0) t1 - t0 t1 - t0 Secant Line: the line determined by two points on a curve

4. Instantaneous Velocity For an object moving along an s-axis, with s= f(t), the instantaneous velocity of the object at time t 0 is: V instantaneous = lim f(t1) – f(t0) t1 - t0 t1 - t0 t1t0t1t0t1t0t1t0 http://www.coolschool.ca/lor/CALC12/un it2/U02L01/averagevelocityvsinstantane ous.swf http://www.coolschool.ca/lor/CALC12/un it2/U02L01/averagevelocityvsinstantane ous.swf

5. Average and Instantaneous Rates of Change Slope can be viewed as a rate of change, and can be useful beyond simple velocity examples. r average = f(x 1 ) – f(x 0 ) x 1 - x 0 x 1 - x 0 If y= f(x), the average rate of change over the interval [x 0, x 1 ] of y with respect to x is: If y= f(x), the instantaneous rate of change of y with respect to x at x 0 is: r inst. = lim f(x1) – f(x0) x1 - x0 x1 - x0 x1x0x1x0x1x0x1x0

6. Examples!! 1: Find the slope of the graph of f(x)= x 2 +1 at the point x 0 = 2 We’re looking for the instantaneous rate of change (slope) of f(x) at x= 2 r inst = lim f(x 1 ) – f(x 0 ) x 1 – x 0 x 1  x 0 r inst = lim f(x 1 ) – f(2) x 1 – 2 x 1  2 r inst = lim (x 1 ) 2 +1 – (2 2 + 1) x 1 – 2 x 1  2 r inst = lim (x 1 – 2) (x 1 + 2) x 1 – 2 x 1  2 r inst = lim (x 1 + 2) x 1  2 r inst = 4

7. Examples!! 2: During the first 40s of a rocket flight, the rocket is propelled straight up so that in t seconds it reaches a height of s=5t 3 ft.  How high does the rocket travel?  What is the average velocity of the rocket during the first 40 sec?  What is the instantaneous velocity of the rocket at the 40 sec mark?

8. Examples!! 2: (cont) How high does the rocket travel? Knowns: s = 5t 3 ft t = 40 sec s= 5 (40) 3 s = 320, 000 ft

9. Examples!! 2: (cont) What is the average velocity of the rocket during the first 40 sec? V average = f(t 1 ) – f(t 0 ) t 1 - t 0 V average = 320000ft – 0 40 sec V average = 8000 ft/sec

10. Examples!! 2: (cont) What is the instantaneous velocity of the rocket at the 40 sec mark? V instantaneous = lim f(t 1 ) – f(t 0 ) t 1 - t 0 t1t0t1t0 V instantaneous = lim 5*(t 1 ) 3 – [5*40 3 ] t 1 – 40 t 1  40 V instantaneous = lim 5*(t 1 ) 3 – 320000 t 1 – 40 t 1  40 V inst = lim 5t 2 +200t + 8000 t 1  40 V inst = 24,000 ft/sec

11. 3.2 The Derivative  Definition of the derivative  Tangent Lines  The Derivative of f with Respect to x  Differentiability  Derivative Notation  Derivatives at the endpoints of an interval

12. Definition of the Derivative The derivative of f at x = x 0 is denoted by f ’(x 0 ) = lim f(x 1 ) – f(x 2 ) x 1 – x 2 x 1 – x 2 x1 x2x1 x2 Assuming this limit exists, f ‘ (x 0 ) = the slope of f at (x 0, f(x 0 )) f ‘ (x 0 ) = the slope of f at (x 0, f(x 0 ))

13. Tangent Lines The tangent line to the graph of f at (x 0, f(x 0 )) is the line whose equation is: y - f(x 0 ) = f‘(x 0 ) * ( x - x 0 )

14. The Derivative of f with Respect to x f ’(x) = lim f(w) – f(x) w – x w – x w  x

15. Differentiability For a given function, if x 0 is not in the domain of f, or if the limit does not exist at x 0, than the function is not differentiable at x 0 The most common instances of nondifferentiability occur at a: NOTE: If f is differeniable at x=x 0, then f must also be continuous at x 0

16. Derivative Notation “the derivative of f(x) with respect to x”

17. Derivatives at the Endpoints of an Interval If a function f is defined on a closed interval [a, b], then the derivative f’(x) is not defined at the endpoints because f ’(x) = lim f(w) – f(x) w – x w – x wxwx is a two-sided limit. is a two-sided limit. Therefore, define the derivatives using one-sided, right and left hand, limits

18. Derivatives at the Endpoints of an Interval A function f is differentiable on intervals [a, b] [a, +∞) (-∞, b] [a, b) (a, b] if f is differentiable at all numbers inside the interval, and at the endpoints (from the left or right)

19. Examples!! 1: Given that f(3) = -1 and f’(x) = 5, find an equation for the tangent line to the graph of y = f(x) at x=3 KNOWNS: F’(x) = slope of the tangent line = 5 Point given = (3, -1) USING POINT SLOPE FORM: y + 1 = (5) (x – 3) y = 5x - 16

20. Examples!! 2: For f(x)=3x 2, find f’(x), and then find the equation of the tangent line to y=f(x) at x = 3 f’(x) = 6a f’(x) = lim f(x) – f(a) x - a xaxa f’(x) = lim 3x 2 – 3a 2 x - a xaxa f’(x) = lim 3x + 3a xaxa KNOWNS: f’(x) = slope of tangent line (6a) point (3, 27) POINT SLOPE FORM: y – 27 = (18) (x – 3) y = 18x - 27

21. 3.3 Techniques of Differeniation  Basic Properties  The Power Rule  The Product Rule  The Quotient Rule

22. Basic Properties

23. The Power Rule

24. The Product Rule The Quotient Rule

25. Examples!! 1: Find dy/dx of y= (x-3) (x 4 + 7) Solve this using the product rule: Let f(x) = (x-3) and g(x)= (x 4 + 7) y’ = 5x 4 - 12x 3 + 7

26. Examples!! Solve this using the quotient rule: Let f(x) = 4x + 1 and g(x) = x 2 - 5 - (4x 2 + 2x + 20) x 4 – 10x 2 + 25 y’ = 4x + 1 x 2 - 5 2: Find dy/dx of y =

27. 3.4 Derivatives of Trigonometric Functions  Derivatives of the Trigonometric Functions (sinx, cosx, tanx, secx, cotx, cscx)

28. Derivatives of Trigonometric Functions!

29. Examples!! sin x sec x 1 + x tan x 1: Find dy/dx of y = Solve this using the quotient and product rules: Simplify. 1 (1 + x tanx) 2 y’ =

30. Examples!! Solve this using the product rule: 2: Find y´ (x) of y = x 3 sin x – 5 cos x dy/dx = x 3 (cos x) + (sin x)(3x 2 ) + 5 sin x

31. 3.5 The Chain Rule  Derivatives of Compositions  The Chain Rule  An Alternate Approach

32. Derivatives of Compositions If you know the derivative of f and g, how can you use these to find the derivative of the composition of f ° g?

33. Chain Rule! If g is differentiable at x and f is differentiable at g(x), then the composition f ° g is differentiable at x If y = f(g(x)) and u = g(x) then y = f(u)

34. An Alternative Approach Sometimes it is easier to write the chain rule as: g(x) is the inside function “The derivative of f(g(x)) is the derivative of the outside function evaluated at the inside function times the derivative of the inside function” f(x) is the outside function

35. An Alternative Approach An Alternative Approach That is:

36. An Alternative Approach Substituting u = g(x) you get

37. Examples!! 1: Find dy/dx of y = (5x + 8) 13 (x 3 + 7x) 12 Use the chain rule, and product rule dy/dx = [(5x +8) 13 ][12(x 3 + 7x) 11 (3x 2 + 7)] + [(x 3 + 7x) 12 ][13(5x + 8) 12 (5)] dy/dx = 12(5x +8) 13 (x 3 + 7x) 11 (3x 2 + 7) + 65(x 3 + 7x) 12 (5x + 8) 12

38. 3.6 Implicit Differentiation  Explicit versus Implicit  Implicit Differentiation

39. Explicit Versus Implicit “A function in the form y = f(x) is said to define y explicitly as a function of x because the variable y appears alone on one side of the equation.” “If a function is defined by an equation in which y is not alone on one side, we say that the function defines y implicitly”

40. Explicit Versus Implicit Implicit: yx + y + 1 = x NOTE: The implicit function can sometimes by rewritten into an explicit function Explicit: y = (x-1) / (x+1)

41. Explicit Versus Implicit “A given equation in x and y defines the function f implicitly if the graph of y = f(x) coincides with a portion of the graph of the equation”

42. Explicit Versus Implicit So, for example the graph of x 2 + y 2 = 1 defines the functions f 1 (x) = √(1-x 2 ) f 2 (x) = -√(1-x 2 ) implicitly, since the graphs of these functions are contained in the circle x 2 + y 2 = 1

43. Explicit Versus Implicit

44. Implicit Differentiation Usually, it is not necessary to solve an equation for y in terms of x in order to differentiate the functions defined implicitly by the equation

45. Examples!! 1: Find dy/dx for sin(x 2 y 2 ) = x cos(x 2 y 2 ) [x 2 (2y) + y 2 (2x)] = 1 dy dx dy dx cos(x 2 y 2 ) 2yx 2 = 1 - 2xy 2 = 1- 2xy 2 cos(x 2 y 2 ) 2yx 2 cos(x 2 y 2 )

46. Examples!! 2: Find d 2 y/dx 2 for x 3 y 3 – 4 = 0 x 3 (3y 2 ) + y 3 (3x 2 ) = 0 dx dy dx dy -y 3 x 2 x 3 y 2 dx dy = x -y = (x)(-1 ) - (-y)(1) dx 2 d2yd2y = x 2 dx dy x 2 x -y dx 2 d2yd2y = -x( ) + y dx 2 d2yd2y = 2y x 2

47. 3.7 Related Rates  Differentiating Equations to Relate Rates

48. Differentiating Equations to Relate Rates Strategy for Solving Related Rates Step 1: “Identify the rates of change that are known and the rate of change that is to be found. Interpret each rate as a derivative of a variable with respect to time, and provide a description of each variable involved”.

49. Differentiating Equations to Relate Rates Strategy for Solving Related Rates Step 2: “Find an equation relating those quantities whose rates are identified in Step 1. In a geometric problem, this is aided by drawing an appropriately labeled figure that illustrates a relationship involving these quantities”.

50. Differentiating Equations to Relate Rates Strategy for Solving Related Rates Step 3: “Obtain an equation involving the rates in Step 1 by differentiating both sides of the equation in Step 2 with respect to the time variable”.

51. Differentiating Equations to Relate Rates Strategy for Solving Related Rates Step 4: “Evaluate the equation found in Step 3 using the known values for the quantities and their rates of change at the moment in question”.

52. Differentiating Equations to Relate Rates Strategy for Solving Related Rates Step 5: “Solve for the value of the remaining rate of change at this moment”.

53. Example!! 1: Sand pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a constant rate of 5ft/ min, at what rate is sand pouring from the chute when the pile is 10 ft high?

54. Example!! STEP1. dt dh = 5 ft/ min = ? dV dt h = 10 ft t = time h = height of conical pile at a given time V = amount of sand in conical pile at a given time 1: (cont.)

55. Example!! STEP2. V = π h 3 12 1: (cont.)

56. Example!! STEP3. V = π h 3 12 1: (cont.) STEP4. dt dV = π h 2 4 dt dh dt dV = π (10ft) 2 4 (5ft/min) = 125π ft 3 / min STEP5.

57. Example!! 2: A 13-ft ladder is leaning against a wall. If the top of the ladder slips down the wall at a rate of 2 ft/sec, how fast will the foot of the ladder be moving away from the wall when the top is 5 ft above the ground?

58. Example!! STEP1. dt dh = -2 ft/ sec = ? dD dt h = 5 ft t = time h = height of the top of the ladder against the wall D = distance of the foot of the ladder from the base of the wall 2: (cont.)

59. Example!! STEP2. 2: (cont.) D 2 + h 2 = 13 2 D 2 + h 2 = 169

60. Example!! STEP3. 2: (cont.) STEP4.STEP5. D 2 + h 2 = 169 = 5/6 ft / sec dt dD 2D + 2h = 0 dt dD dt dh 24ft + 10ft (-2 ft/sec) = 0 dt dD (note: at h=5, D = 12)

61. 3.8 Local Linear Approximation; Differentials  Local Linear Approximation  Differentials

62. Local Linear Approximation “Linear Approximation may be described informally in terms of the behavior of the graph of f under magnification: if f is differentiable at x 0, then stronger and stronger magnifications at a point, P, eventually make the curve segment containing P look more and more like a nonvertical line segment, that line being the tangent line to the graph of f at P.”

63. Local Linear Approximation A function that is differentiable at x 0 is said to be locally linear at the point P (x 0, f(x 0 )) As you zoom closer to a point P, the function looks more and more linear

64. Local Linear Approximation Assume that a function f is differentiable at x 0, and remember that the equation of the tangent line at the point P (x 0, f(x 0 )) is:

65. Local Linear Approximation Since the tangent line closely approximates the graph of f for values of x near x 0, that means that provided x is near x 0, then: This is called the local linear approximation of f at x 0

66. Local Linear Approximation By rewriting the formula with ∆x = x - x 0, you get:

67. Local Linear Approximation Generally, the accuracy of the local linear approximation to f(x) at x 0 will deteriorate as c gets progressively farther from x 0.

68. Differentials Early in the development of calculus, the symbols “dy” and “dx” represented “infinitely small changes” in the variables y and x. The derivative dy/dx was thought to be a ratio of these infinitely small changes. However, the precise meaning is logically elusive. Our goal is to define the symbols dy and dx so that dy/dx can actually be treated as a ratio Our goal is to define the symbols dy and dx so that dy/dx can actually be treated as a ratio

69. Differentials The variable dx is called the differential of x. If we are given a function y = f(x) tha is differentiable at x = x 0, then we define the differential of f at x 0 to be the function of dx given by the formula:

70. Differentials The symbol dy is simply the dependant variable of this function, and is called the differential of y. dy is proportional to dx with constant of proportionality f ‘ (x 0 ). If dx is not 0, you can obtain

71. Differentials Because f ‘ (x) is equal to the slope of the tangent line to the graph of f at the point (x,f(x)), the differentials dy and dx can be described as the rise and run of this tangent line.

72. EXAMPLES!! 1: Find the local linear approximation of x 3 at x 0 = 1 y ≈ 1 + 3(x-1) y ≈ 3x + 2

73. EXAMPLES!! 2: Use an appropriate local linear approximation to estimate the value of √ (36.03) f(36.03) ≈ f(36) + (1/12)(36.03-36) f(36.06) ≈ 6.0025 Let f(x)=√(x) f ‘ (x)= (1/2)x -(1/2) f(36.03) ≈ 6 + (1/12)(.03)

74. EXAMPLES!! 3: Find the differential dy for y = xcosx dy/dx = -xsinx + cosx dy = (-xsinx + cosx) dx

75. BIBLIOGRAPHY!!!  http://mathcs.holycross.edu/~spl/old_courses/131_fall_2005/tangent_li ne.gif  http://www.coolschool.ca/lor/CALC12/unit3/U03L08/example_07.gif  http://www.clas.ucsb.edu/staff/Lee/Tangent%20and%20Derivative.gif  http://images.search.yahoo.com/search/images/view?back=http%3A% 2F%2Fimages.search.yahoo.com%2Fsearch%2Fimages%3Fp%3Dpro duct%2Brule%2Bcalculus%26ei%3DUTF-8%26fr%3Dyfp-t- 501%26x%3Dwrt&w=309&h=272&imgurl=www.karlscalculus.org%2Fq rule_still.gif&rurl=http%3A%2F%2Fwww.karlscalculus.org%2Fcalc4_3. html&size=15.1kB&name=qrule_still.gif&p=product+rule+calculus&type =gif&no=4&tt=53&oid=0077160168cbaf58&ei=UTF-8