1 / 75

Chapter 3

Chapter 3. The Derivative. By: Kristen Whaley. 3.1 Slopes and Rates of Change. Average Velocity Instantaneous Velocity Average Rate of Change Instantaneous Rate of Change. V average = f(t1) – f(t0) t1 - t0. Average Velocity.

gurley
Download Presentation

Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 The Derivative By: Kristen Whaley

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

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

  4. V instantaneous = lim f(t1) – f(t0) t1 - t0 t1t0 Instantaneous Velocity For an object moving along an s-axis, with s= f(t), the instantaneous velocity of the object at time t0 is: http://www.coolschool.ca/lor/CALC12/unit2/U02L01/averagevelocityvsinstantaneous.swf

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

  6. r inst = lim f(x1) – f(x0) x1 – x0 x1 x0 r inst = lim f(x1) – f(2) x1 – 2 r inst = lim (x1)2 +1 – (22 + 1) x1 – 2 r inst = lim (x1 + 2) x1 2 x1 2 x1 2 r inst = lim (x1 – 2) (x1 + 2) x1 – 2 r inst = 4 x1 2 Examples!! 1: Find the slope of the graph of f(x)= x2+1 at the point x0 = 2 We’re looking for the instantaneous rate of change (slope) of f(x) at x= 2

  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=5t3 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. s = 320, 000 ft Examples!! 2: (cont) How high does the rocket travel? Knowns: s = 5t3 ft t = 40 sec s= 5 (40)3

  9. V average = f(t1) – f(t0) t1 - t0 V average = 320000ft – 0 40 sec V average = 8000 ft/sec Examples!! 2: (cont) What is the average velocity of the rocket during the first 40 sec?

  10. V inst= lim5t2 +200t + 8000 t140 V instantaneous = lim f(t1) – f(t0) t1 - t0 V instantaneous = lim 5*(t1)3 – [5*403] t1 – 40 V instantaneous = lim 5*(t1)3 – 320000 t1 – 40 t1t0 V inst = 24,000 ft/sec t140 t140 Examples!! 2: (cont) What is the instantaneous velocity of the rocket at the 40 sec mark?

  11. 3.2The 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 = x0 is denoted by f ’(x0) = lim f(x1) – f(x2) x1 – x2 x1x2 Assuming this limit exists, f ‘ (x0) = the slope of f at (x0, f(x0))

  13. Tangent Lines The tangent line to the graph of f at (x0, f(x0)) is the line whose equation is: y - f(x0) = f‘(x0) * ( x - x0 )

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

  15. Differentiability For a given function, if x0 is not in the domain of f, or if the limit does not exist at x0, than the function is not differentiable at x0 NOTE: If f is differeniable at x=x0, then f must also be continuous at x0 The most common instances of nondifferentiability occur at a:

  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 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. y = 5x - 16 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)

  20. f’(x) = lim f(x) – f(a) x - a f’(x) = lim 3x2 – 3a2 x - a xa xa f’(x) = lim 3x + 3a xa f’(x) = 6a y = 18x - 27 Examples!! 2: For f(x)=3x2 , find f’(x), and then find the equation of the tangent line to y=f(x) at x = 3 KNOWNS: f’(x) = slope of tangent line (6a) point (3, 27) POINT SLOPE FORM: y – 27 = (18) (x – 3)

  21. 3.3Techniques 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. Solve this using the product rule: y’ = 5x4 - 12x3 + 7 Examples!! 1: Find dy/dx of y= (x-3) (x4 + 7) Let f(x) = (x-3)and g(x)= (x4 + 7)

  26. 4x + 1 2: Find dy/dx of y = x2 - 5 Solve this using the quotient rule: - (4x2 + 2x + 20) y’ = x4 – 10x2 + 25 Examples!! Let f(x) = 4x + 1 and g(x) = x2 - 5

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

  28. Derivatives of Trigonometric Functions!

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

  30. dy/dx = x3 (cos x) + (sin x)(3x2) + 5 sin x Examples!! 2: Find y´ (x) of y = x3sin x – 5 cos x Solve this using the product rule:

  31. 3.5The 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 f(x) is the outside 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”

  35. An Alternative Approach That is:

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

  37. dy/dx = 12(5x +8)13(x3 + 7x)11(3x2 + 7) + 65(x3 + 7x)12 (5x + 8)12 Examples!! 1: Find dy/dx of y = (5x + 8)13(x3 + 7x)12 Use the chain rule, and product rule dy/dx = [(5x +8)13][12(x3 + 7x)11(3x2 + 7)] + [(x3 + 7x)12][13(5x + 8)12(5)]

  38. 3.6Implicit 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 x2 + y2 = 1 defines the functions f1(x) = √(1-x2) f2(x) = -√(1-x2) implicitly, since the graphs of these functions are contained in the circle x2 + y2 = 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. cos(x2y2) [x2(2y) + y2 (2x)] = 1 = 1 - 2xy2 dy dy cos(x2y2) = 1- 2xy2cos(x2y2) dx dx 2yx2cos(x2y2) 2yx2 Examples!! 1: Find dy/dx for sin(x2y2) = x

  46. x3(3y2) + y3(3x2) = 0 -y3x2 -y -y x2 = = x x3y2 d2y dy d2y d2y dy dy dy dx dx2 dx dx2 dx2 dx dx (x)(-1 ) - (-y)(1) = x2 2y -x( ) + y = = x2 x Examples!! 2: Find d2y/dx2 for x3y3 – 4 = 0

  47. 3.7Related 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”.

More Related