2 Average Rate of Change of f over [a, b]: Difference Quotient The average rate of change of the function f over the interval [a, b] is Average rate of change of f = f = f(b) - f(a) =Slope of line through points P and Q in the figure x b - a

3 This average rate of change the diff quotient of f over the interval [a, b].

4 Ex: Let f(x) = x 3 + x. Then: Rate of change of f over [2, 4]= f(4) - f(2) = 68-10/2=29 Rate of change of f over [a, a+h]= f(a+h) - f(a) = 3a 2 + 3ah +h (a+h) - a

5 A Numerical Approach In Indonesia, you monitor the value of the US Dollar on the foreign exchange market very closely during a rather active five-day period. Suppose R(t) = 7, t - 100t 2 rupiahs, The rupiah is the Indonesian currency, where t is time in days. (t = 0 represents the value of the Dollar at noon on Monday.)

6 What was the value of the Dollar at noon on Tuesday? According to the graph, when was the value of the Dollar rising most rapidly?

7 A formula for the average rate of change of the Dollar's value over the interval [1, 1+h] is given by Use your answer to the last question to complete the following table.

8 Instantaneous Rate of Change of f(x) at x = a: The Derivative The instantaneous rate of change of f(x) at x = a is defined by taking the limit of the average rates of change of f over the intervals [a, a+h], as h approaches 0.

9 The instantaneous rate of change the derivative of f at x = a which we write as f'(a).

10 The Derivative as Slope: A Geometric Approach Estimating the Slope by Zooming In

11 Notice how the curve appears to "flatten" as we zoom in; Before Zooming InAfter Zooming In

12 Slope of the Secant Line and Slope of the Tangent Line The slope of the secant line through (x, f(x)) and (x+h, f(x+h)) is the same as the average rate of change of f over the interval [x, x+h], or the difference quotient:

13 The slope of the tangent line through (x, f(x)) is the same as the instantaneous rate of change of f at the point x, or the derivative:

14 Ex: Let f(x) = 3x 2 + 4x. Use a difference quotient with h = to estimate the slope of the tangent line to the graph of f at the point where x = 2. Sol:

15 The Derivative as a Function: An Algebraic Approach So far, all we have been doing is approximating the derivative of a function. Is there a way of computing it exactly?

16 Recall: The derivative of the function f at the point x is the slope of the tangent line through (x, f(x)), or the instantaneous rate of change of f at the point x.

17 The slope of the tangent, or derivative, depends on the position of the point P on the curve, and therefore on the choice of x.

18 Therefore, the derivative is a function of x, and that is why we write it as f'(x) f'(1) = slope of the tangent at the point on the graph where x = 1. f'(-4) = slope of the tangent at the point on the graph where x = -4.

19 Definition The derivative f'(x) of the function f(x) is the slope of the tangent at the point (x, f(x)).

20 In words, the derivative is the limit of the difference quotient. By the "difference quotient" we mean the average rate of change of f over the interval [x, x+h]:

21 Definition A derivative f'(x) of a function f depicts how the function f is changing at point x. f must be continuous at point x in order for there to be a derivative at that point. A function which has a derivative is said to be differentiable.

22 The derivative is computed by using the concept of x. x is an arbitrary change or increment in the value of x.

23 Ex: Let f(x) = 3x 2 + 4x. The difference quotient is given by: Hint: Average rate of change of f over [x, x+h] = Now take the limit as h 0. Ex Continued : f'(x) = f'(1) =

24 Ex: Let f(x) = 1/x, f(x+h) is given by The difference quotient is given by:

25 Power Rule

26 Negative Exponents Since the power rule works for negative exponents, we have, for

27 Ex: If f(x) = x 3, then f'(x) = 3x 2. When we say "f'(x) = 3x 2," "The derivative of x 3 with respect to x equals 3x 2." “The derivative with respect to x" by the symbol "d/dx."

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29 Derivatives of Sums, Differences & Constant Multiples

30 The quotient f(x)/g(x) If f(x) and g(x) are diff. Then d (f(x)/g(x)) = lim f(x+h)/g(x+h) -(f(x)/g(x)) = lim f(x+h).g(x) - f(x).g(x+h) = lim f(x+h).g(x) -f(x)g(x) +f(x)g(x) - f(x).g(x+h) = lim g(x). (f(x+h)-f(x))/h - f(x). (g(x+h)-g(x))/h dx h 0 h h g(x)g(x+h) h 0 h g(x)g(x+h) h 0 g(x)g(x+h)

31 dx g(x).f'(x) - f(x).g'(x)d (f(x)/g(x)) = g(x).g(x)

32 Ex:

33 Limits and Continuity: Numerical Approach Estimating Limits Numerically "What happens to f(x) as x approaches 2?" Calculating the limit of f(x) as x approaches 2,

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35 Ex: lim g(x) = -5 x - lim g(x) = -5 x lim g(x) = -5 x + Ex: lim f(x) = ? x 3

36 Limits and Continuity: Graphical Approach Estimate x -2 f(x)

37 To decide whether x a f(x) exists, and to find its value if it does. 1.Draw the graph of f(x) either by hand or using a graphing calculator. 2.Position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the right of x = a. In the example illustrated, we are estimating lim x -2 f(x)

38 3. Move the point along the graph toward x = a from the right. The value the y-coordinate approaches (if any) is lim x a+a+ f(x)

39 The y-coordinate is approaching 2 as x approaches - 2 from the right. Therefore, lim x -2 + f(x)=2

40 Repeat Steps 2 and 3, but this time starting from a point on the graph to the left of x = a, and approach x = a along the graph from the left. The y-coordinate approaches (if any) is then lim x a-a- f(x)=

41 The y-coordinate is again approaching 2 as x approaches -2 from the left. lim x -2 - f(x)=2

42 5. If the left and right limits both exist and have the same value L, then lim f(x) = exists and equals L. The left and right limits both exist and equal 2, and so x a lim x -2 f(x)=2

43 Limits and Continuity: Algebraic Approach lim x 2 X 2 -3x 2x+3 = …=-2/11 notice that you can obtain the same answer by simply substituting x = 2 in the given function: f(x)= X 2 -3x 2x …=-2/11f(2)=

44 Is that all there is to evaluating limits algebraically: just substitute the number x is approaching in the given expression?

45 Ans: The function is continuous at the value of x in question.

46 Continuous Functions The function f(x) is continuous at x=a if lim f(x) exists and equals f(a). The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular a, we say that f is discontinuous at a or that f has a discontinuous at a. x a

47 Let us evaluate lim 3x 2 +x-10 x -2 x+2 Ask yourself the following questions: Is the function f(x) a closed form function? Is the value x = a in the domain of f(x)? -11

48 The statement d dx (x 2 /x 5 )=2x/5x 4 is: Wrong, because the correct answer is (a) -3/x 4 (b) 0/3x 2 =0 (c) 1/3x 2 (d) lnx 3

49 Quadratic formula - derivation For quadratic equations of the type x 2 + p x + q = 0 The derivation of the quadratic formula for the roots of ax 2 +bx+c=0.

50 We are going to solve for x. ax 2 +bx+c=0 Divided through by a. x 2 + b/a x+ c/a=0 Subtracted c/a on both sides. x 2 + b/a x = -c/a Complete the square on the left. x 2 + b/a x + (b/2a) 2 = -c/a + (b/2a) 2 The left is square (x+ b/2a) 2 = -c/a + (b/2a) 2

51 Common dominator is 4a 2 (x+ b/2a) 2 = -4ac+b 2 /4a 2 Now take the square roots. (x+ b/2a) 2 = b 2 -4ac (x+ b/2a) = √ Subtract b/2a on both sides X = - b/2a √ 4a 2 + b 2 -4ac 2a + b 2 -4ac 2a

52 The Product Rule

53 The derivative of a product is NOT the product of the derivatives.

54 In the form of u(x+h)v(x) -u(x+h)v(x) to the numerator

55 In the last step, because u(x) is differentiable at x and therefore continuous. The product u(x)v(x) as the area of a rectangle with width u(x) and height v(x). The change in area is d(uv), and is indicated is the figure below.

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57 As x changes, the area changes from the area of the red rectangle, u(x)v(x), to the area of the largest rectangle, the sum of the read, green, blue and yellow rectangles. The change in area is the sum of the areas of the green, blue and yellow rectangles,

58 Product and Quotient Rule Product Rule Quotient Rule

59 Ex: Find the derivative of f(x) =(4x 3 -x 4 )(11x-√x). Sol: First recognize that f(x) is a product of two factors: (4x 3 -x 4 ) and (11x-√x) Rewrite the function in exponent form f(X)= (4x 3 -x 4 )(11x-x 0.5 ) d dx (4x 3 -x 4 )(11x-x 0.5 ) =(12x 2 -4x 3 ) (11x-x 0.5 ) +(4x 3 -x 4 ) (11-0.5x -0.5 )

60 In the limit of dx small, the area of the yellow rectangle is neglected. Algebraically,

61 Ex: For c is a constant,

62 Ex: Whether or not this is substantially easier than multiplying out the polynomial and differentiating directly is a matter of opinion; decide for yourself.

63 Ex: If f and g are differentiable functions such that f(2)=3, f’(2)=-1, g(2)=-5 and g’(2)=2, then what is the value of (fg)’(2)? Ex: With g(x)=(x 3 -1)(x 3 +1) what is g ’(x)? EX: Find dy/dx where y(x)= (8x-1) (x 2 +4x+7)(x 3 -5)

64 Ex: If f, g and h are differentiable, use the product rule to show that As a corollary, show that

65 Ex: d dx (x 2 -x)(5+x -0.5 ) Ex: d dx (x 2 -1/x 2 )(5+x -0.5 )

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67 Ex: The derivative of f(x) = x 2 +3x+2 is? x-1 Using the Calculation Thought Experiment (CTE) Let us use the CTE to find the derivative of f(x)=(3x+1) x 2 +4 x 2 +x

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69 Ex: Find out the derivative of (3x /x)(x + 1) Ex: Find out the derivative of f(x)=4x 2 +(x-1)(4x+1) 3x

70 Suppose we want to find the derivative of y(x) = (x 2 +3x+1) 2 We could hopefully multiply y(x) out and then take the derivative with little difficulty. But, what if, y(x) = (x 2 +3x+1) 50 Would you want to apply the same method to this problem?

71 Ex: Returning to the first y(x) above, if we let The function y(x) is the composite of g with f.

72 Our goal is to find the derivative Based on our knowledge of the functions f and g. Now, we know that Leading to the speculation that

73 This leads to the (possible) chain rule: Ex:

74 Ex: The function sin(2x) is the composite of the functions sin(u) and u=2x. Then, Ex:

75 Ex: sin2(4x) is a composite of three functions : u 2, u=sin(v) and v=4x. As a check, you may want to note that the above may be expressed as

76 Inverse Function: To find the derivative based on the knowledge or condition that for some function f(t), or, in other words, that g(x) is the inverse of f(t) = x. Recognizing that t and g(x) represent the same quantity, and remembering the Chain Rule,

77 This result becomes somewhat obvious;

78 Ex: We know from the Power Rule, with n=2, that Equivalently

79 The above may be generalized; for nonzero n, Then

80 Rule name (if any) The Sum rule The Product rule The Quotient rule

81 The Chain rule The Power rule

82 Derivatives of Hyperbolic Functions:

83 Taking f(x)= x 3 we get d dx U 3 = 3u 2 d dx u The Chain Rule If u is a diff. function of x, and f is a diff. function of u, then: d dx [f(u)]= f’(u) d dx u

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85 Ex: d dx [ ] 5 = X 2 -x -1 3x-1 [(x 2 - 1) 3 (3x + 4) -1 ]= ? d dx Ex:

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87 Derivatives of Logarithms: Let u be a function of x, Derivatives of Exponential Functions: Let u be a function of x,

88 Derivatives of Inverse Hyperbolic Functions: If u is a function of x;

89 Implicit Differentiation Not all functions are given explicitly and are only implied by an equation. Ex: xy = 1 is an equation given implicitly, explicitly it is y = 1/ x. Now to find dy/dx for xy = 1, simply solve for y and differentiate. xy = 1 y = 1 / x = x -1 dy/dx = -1 x -2 = 1/x 2

90 But, not all equations are easily solved for y, as in the equation 3x + y 3 = y This is where implicit differentiation is applied. Implicit differentiation is taking the derivative of both sides of the equation with respect to one of the variables. Most commonly, used is the derivative of y with respect to x. or dy/dx.

91 Ex: 3x + y 3 = y 2 + 4, solve for dy/dx. 3x + y 3 = y d/dx(3x + y 3 ) = d/dx(y 2 + 4) 3 + 3y 2 dy/dx = 2y dy/dx 3 = 2y dy/dx - 3y 2 dy/dx 3 = y ( 2 - 3y ) dy/dx 3 / y (2 - 3y ) = dy/dx

92 Ex: Find the slope of the curve x 2 + y 3 = 2x + y at ( 2,4) Sol: d/dx [x 2 + y 3 ] = d/dx [2x + y] 2x + 3y 2 dy/dx = 2 + dy/dx 2x - 2 = (-3y 2 + 1) dy/dx 2( x - 1) / (-3y 2 + 1) = dy/dx = slope of curve substitute (2,4) into dy/dx to find the slope at that point. 2(2-1) / (-3 · ) = 2 /-49 = -2/49 is the slope of the curve.

93 Derivatives of Higher Order Derivatives of functions are also functions, therefore can be differentiated again. Ex: f(x) = x 5 f '(x) = 5x 4 f ''(x) = 5·4x 3 = 20x 3 f '''(x) = 5·4·3x 2 = 60x 2

94 Maximum and Minimum Values of a Function Second Derivative Test for Functions Concavity: If the second derivative of a function f ( f ''(x) ) is positive (or negative) for all x on (a,b) then the graph of f is concave upward (or downward) on (a,b).

95 Second Derivative Test for Max. and Min. Points. If point A(a, f (a)) is on the graph of function f such that f '(a) = 0 and f ''(a) < 0, then point A is a relative maximum; if f '(a) = 0 and f ''(a) > 0, then point A is a relative minimum.

96 Maximum and Minimum Values of a Function Increasing and Decreasing Functions: A function f is said to be increasing when f '(x) > 0 for every x on (a,b) and decreasing when f '(x) < 0 for every x on (a,b). Absolute Max. and Min. of a function: The absolute maximum (or minimum) of a function is a point (a, f (a)) when f (a)  f (x) ( or f (a) < f (x) ) for any value of x in the domain of f.

97 Ex: Find the open intervals on which the function f (x) = x x 2 is increasing or decreasing. Sol: f '(x) = 3 x x let f '(x) = 3 x x = 0 x = 0 or 2 Critical numbers Because there are no x-values for which f ' is undefined, it follows that x = 0 and x = 2 are the only critical numbers. So, the intervals that need to be tested are (-, 0), (0, 2), and (2, ).

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109 Critical Point: A critical point, (x, f (x)), of a function f is if f (x) is defined and f '(x) is either zero or undefined. The x-coordinate of the critical point is called a critical value or a critical number.

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