February 21 Math 1190 sec. 63 Spring 2017

Transcription

1 February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative of f at c is provie the limit exists. f f (x) f (c) f (c + h) f (c) (c) = lim = lim x c x c h 0 h February 17, / 105

2 Notes on Notation If y = f (x), we have the Leibniz notation for the erivative function f (x) = y x. We can use this notation for the erivative at a point in the following well accepte way f (c) = y x. c For example, if y = x 2, then y x = 2x, an y x = 6. 3 February 17, / 105

3 The Derivative Function For y = f (x), the erivative y x = f f (x + h) f (x) (x) = lim. h 0 h Another, sometimes convenient formulation of this is y x = f f (z) f (x) (x) = lim z x z x using a ummy variable (a place keeper) z. February 17, / 105

4 Derivative Rules Constant: x c = 0 Ientity: x x = 1 Power: x x n = nx n 1 for integers n Exponential: x ex = e x February 17, / 105

5 Derivative Rules Constant Factor: x kf (x) = kf (x) Sum or Difference: x (f (x)±g(x)) = f (x)±g (x) Prouct: x f (x)g(x) = f (x)g(x)+f (x)g (x) Quotient: f (x) x g(x) = f (x)g(x) f (x)g (x) (g(x)) 2 February 17, / 105

6 Six Trig Function Derivatives sin x = cos x, x cos x = sin x, x x tan x = sec2 x, x cot x = csc2 x, sec x = sec x tan x, x csc x = csc x cot x x February 17, / 105

7 Question Suppose y = 2xe x. Evaluate y x 0 (a) (b) (c) () y x = 0 0 y x = 2 0 y x = 2e 0 y x = e 0 February 17, / 105

8 Question True or False: If f is ifferentiable at c, then the slope, m tan, of the line tangent to the graph of f at the point (c, f (c)) is m tan = f (c). February 17, / 105

9 Question Using the quotient rule, x cos x x = (a) (b) (c) () sin x 3x 2 (x 3 + 2) sin x 3x 2 cos x x (x 3 + 2) sin x + 3x 2 cos x (x 3 + 2) 2 (x 3 + 2) sin x 3x 2 cos x (x 3 + 2) 2 February 17, / 105

10 Section 3.1: The Chain Rule Suppose we wish to fin the erivative of f (x) = (x 2 + 2) 2. February 17, / 105

11 Compositions Now suppose we want to ifferentiate G(x) = (x 2 + 2) 10. How about F(x) = x 2 + 2? February 17, / 105

12 Example of Compositions Fin functions f (u) an g(x) such that F(x) = x = (f g)(x). February 17, / 105

13 Example of Compositions Fin functions f (u) an g(x) such that ( πx ) F(x) = cos = (f g)(x). 2 February 17, / 105

14 Theorem: Chain Rule Suppose g is ifferentiable at x an f is ifferentiable at g(x). Then the composite function F = f g is ifferentiable at x an x F (x) = x f (g(x)) = f (g(x))g (x) In Liebniz notation: if y = f (u) an u = g(x), then y x = y u u x. February 17, / 105

15 Example Determine any insie an outsie functions an fin the erivative. (a) F(x) = sin 2 x February 17, / 105

16 (b) F(x) = e x 4 5x 2 +1 February 17, / 105

17 Question If ( πθ G(θ) = cos 2 π ) = f (g(θ)) where 4 f (u) = cos u, an g(θ) = πθ 2 π 4, then (a) (b) (c) G (θ) = π 2 sin θ G (θ) = π ( πθ 2 sin 2 π ) 4 ( ) πθ ( π ) G (θ) = sin +sin 2 4 February 17, / 105

18 The power rule with the chain rule If u = g(x) is a ifferentiable function an n is any integer, then x un n 1 u = nu x. Evaluate: x e7x = x (ex ) 7 February 17, / 105

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20 Question Evaluate x sin4 (x) = x (sin x)4 (a) 4 cos 3 (x) (b) 4 sin 3 (x) cos 3 (x) (c) 4 sin 3 (x) cos(x) () 4 sin 3 (x) cos(x) February 17, / 105

21 Example Fin the equation of the line tangent to the graph of y = cos 2 x at the point ( π 4, 1 2 ). February 17, / 105

22 Figure: y = cos 2 x an the tangent line at ( π 4, 1 2 ). February 17, / 105

23 We may be able to choose between ifferentiation methos. Evaluate x sin x x using (a) The quotient rule: February 17, / 105

24 (b) writing sin x x = (sin x)(x 3 +2) 1 an using the chain rule. February 17, / 105

25 Question Suppose that g(x) is ifferentiable. The erivative of e g(x) is (Hint: The outsie function is e u, an the insie function is g(x).) (a) e g(x) g (x) (b) e g(x) (c) e g (x) February 17, / 105

26 Question Fin the equation of the line tangent to the graph of f (x) = e sin x at the point (0, f (0)). (a) y = x + 1 (b) y = 1 (c) y = x 1 () y = ex + 1 February 17, / 105

27 Multiple Compositions The chain rule can be iterate to account for multiple compositions. For example, suppose f, g, h are appropriately ifferentiable, then x (f g h)(x) = x f (g(h(x)) = f (g(h(x)) g (h(x)) h (x) Note that the outermost function is f, an its inner function is a composition g(h(x)). So the erivative of the outer function evaluate at the inner is f (g(h(x)) which is multiplie by the erivative of the inner function itself base on the chain rule g (h(x))h (x). February 17, / 105

28 Example Evaluate the erivative ( ) 1 t tan2 3 t3 February 17, / 105

29 Exponential of Base a Let a > 0 with a 1. By properties of logs an exponentials a x = e (ln a)x. Theorem: (Derivative of y = a x ) Let a > 0 an a 1. Then x ax = a x ln a February 17, / 105

30 Exponential of Base a Theorem: (Derivative of y = a x ) Let a > 0 an a 1. Then x ax = a x ln a Recall from before that if f (x) = a x then x ax = f (0)a x where f a h 1 (0) = lim. h 0 h We now see that the number f (0) = ln(a). February 17, / 105

31 Example Evaluate (a) x 4x (b) x 2cos x February 17, / 105

32 Question x 7x 2 = (a) 7 x 2 (b) (2x)7 x 2 (c) (2x)7 x 2 ln 7 () 7 x 2 ln 7 February 17, / 105

33 Section 3.2: Implicit Differentiation; Derivatives of the Inverse Trigonometric Functions The chain rule states that for a ifferentiable composition f (g(x)) x f (g(x)) = f (g(x))g (x) For y = f (u) an u = g(x) y x = y u u x February 17, / 105

34 Example Assume f is a ifferentiable function of x. Fin an expression for the erivative: x (f (x))2 x tan (f (x)) February 17, / 105

35 Example Suppose we know that y = f (x) for some ifferentiable function (but we on t know exactly what f is). Fin an expression for the erivative. x y 2 x x 2 y 2 February 17, / 105

36 Question If y is some function of x (but we on t know the specifics), then x y 3 = (a) ( ) y 3 x (b) 3y 2 y x (c) 3 ( ) y 2 x February 17, / 105

37 Let s Double Check with an Example: Let y = x 2 so that y x = 2x an y 3 = x 6. Compute with the power rule x y 3 = x x 6 Now take a moment an compute each of ( ) y 3 (a) x (b) 3y 2 y x (c) 3 ( ) y 2 x February 17, / 105

38 Implicitly efine functions A relation an equation involving two variables x an y such as x 2 + y 2 = 16 or (x 2 + y 2 ) 3 = x 2 implies that y is efine to be one or more functions of x. February 17, / 105

39 Figure: x 2 + y 2 = 16 February 17, / 105

40 Figure: (x 2 + y 2 ) 3 = x 2 February 17, / 105

41 Explicit -vs- Implicit A function is efine explicitly when given in the form y = f (x). A function is efine implicitly when it is given as a relation for constant C. F(x, y) = C, February 17, / 105

42 Implicit Differentiation Since x 2 + y 2 = 16 implies that y is a function of x, we can consier it s erivative. Fin y x given x 2 + y 2 = 16. February 17, / 105

43 Show that the same result is obtaine knowing y = 16 x 2 or y = 16 x 2. February 17, / 105

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45 Example Fin y x given x 2 3xy + y 2 = y. February 17, / 105

46 Fining a Derivative Using Implicit Differentiation: Take the erivative of both sies of an equation with respect to the inepenent variable. Use all necessary rules for ifferenting powers, proucts, quotients, trig functions, exponentials, compositions, etc. Remember the chain rule for each term involving the epenent variable (e.g. mult. by y x as require). Use necessary algebra to isolate the esire erivative y x. February 17, / 105

47 Example Fin S r. e Sr + S = r February 17, / 105

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49 Example Fin the equation of the line tangent to the graph of x 3 + y 3 = 6xy at the point (3, 3). February 17, / 105

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52 Figure: Folium of Descartes x 3 + y 3 = 6xy February 17, / 105

53 Figure: Folium of Descartes with tangent line at (3, 3) February 17, / 105

54 Question Fin the equation of the line tangent to the graph of y 2 + y + x 2 = 6 at the point (2, 1). (a) y = 6x + 13 (b) y = x + 3 (c) y = 4 3 x () y = 5 2 x + 6 February 17, / 105

55 The Power Rule: Rational Exponents Let y = x p/q where p an q are integers. This can be written implicitly as y q = x p. Fin y x. February 17, / 105

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57 The Power Rule: Rational Exponents Theorem: If r is any rational number, then when x r is efine, the function y = x r is ifferentiable an for all x such that x r 1 is efine. x x r = rx r 1 February 17, / 105

58 Examples Evaluate (a) x 4 x (b) v csc( v) February 17, / 105

59 Question Fin g (t) where g(t) = 5 t. (a) g (t) = 5t 4 (b) g (t) = 5t 6 (c) g (t) = 1 5t 4/5 () g (t) = 1 5 t4/5 February 17, / 105

60 Inverse Functions Suppose y = f (x) an x = g(y) are inverse functions i.e. (g f )(x) = g(f (x)) = x for all x in the omain of f. Theorem: Let f be ifferentiable on an open interval containing the number x 0. If f (x 0 ) 0, then g is ifferentiable at y 0 = f (x 0 ). Moreover y g(y 0) = g (y 0 ) = 1 f (x 0 ). Note that this refers to a pair (x 0, y 0 ) on the graph of f i.e. (y 0, x 0 ) on the graph of g. The slope of the curve of f at this point is the reciprocal of the slope of the curve of g at the associate point. February 17, / 105

61 Example The function f (x) = x 7 + x + 1 has an inverse function g. Determine g (3). February 17, / 105

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63 Inverse Trigonometric Functions Recall the efinitions of the inverse trigonometric functions. y = sin 1 x x = sin y, 1 x 1, π 2 y π 2 y = cos 1 x x = cos y, 1 x 1, 0 y π y = tan 1 x x = tan y, < x <, π 2 < y < π 2 February 17, / 105

64 Inverse Trigonometric Functions There are ifferent conventions use for the ranges of the remaining functions. Sullivan an Mirana use y = cot 1 x x = cot y, < x <, 0 < y < π y = csc 1 x x = csc y, x 1, y ( π, π ] ( 0, π ] 2 2 y = sec 1 x x = sec y, x 1, y [ 0, π ) [ π, 3π 2 2 ) February 17, / 105

65 Derivative of the Inverse Sine Use implicit ifferentiation to fin x sin 1 x, an etermine the interval over which y = sin 1 x is ifferentiable. February 17, / 105

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68 Examples Evaluate each erivative (a) x sin 1 (e x ) February 17, / 105

69 Examples Evaluate each erivative ( ) 3 (b) sin 1 x x February 17, / 105

70 Derivative of the Inverse Tangent Theorem: If f (x) = tan 1 x, then f is ifferentiable for all real x an f (x) = x tan 1 x = x 2. The argument uses implicit ifferentiation just like we use for the inverse sine function. It is left as an EXERCISE. February 17, / 105

71 Question: u tan 1 u = 1 1+u 2 Fin y x where y = tan 1 (e x ). (a) (b) (c) () y x = y x = ex 1 + e 2x ex 1 + x 2 y x = e 2x y x = sec 2 (e x ) February 17, / 105

72 Derivative of the Inverse Secant Theorem: If f (x) = sec 1 x, then f is ifferentiable for all x > 1 an f (x) = x sec 1 x = 1 x x 2 1. February 17, / 105

73 Examples Evaluate (a) x sec 1 (x 2 ) (b) x tan 1 (sec x) February 17, / 105

74 The Remaining Inverse Functions Due to the trigonometric cofunction ientities, it can be shown that cos 1 x = π 2 sin 1 x an cot 1 x = π 2 tan 1 x csc 1 x = π 2 sec 1 x February 17, / 105

75 Derivatives of Inverse Trig Functions x sin 1 x = 1 1 x 2, 1 x cos 1 x = 1 x 2 x tan 1 x = x 2, x cot 1 x = x 2 x sec 1 x = 1 x x 2 1, 1 x csc 1 x = x x 2 1 February 17, / 105

76 Section 3.3: Derivatives of Logarithmic Functions Recall: If a > 0 an a 1, we enote the base a logarithm of x by log a x This is the inverse function of the (one to one) function y = a x. So we can efine log a x by the statement y = log a x if an only if x = a y. Our present goal is to use our knowlege of the erivative of an exponential function, along with the chain rule, to come up with a erivative rule for logarithmic functions. February 17, / 105

77 Properties of Logarithms We recall several useful properties of logarithms. Let a, b, x, y be positive real numbers with a 1 an b 1, an let r be any real number. log a (xy) = log a (x) + log a (y) ( ) log x a y = log a (x) log a (y) log a (x r ) = r log a (x) log a (x) = log b (x) log b (a) (the change of base formula) log a (1) = 0 February 17, / 105

78 Question (1) In the expression ln(x), what is the base? (a) 10 (b) 1 (c) e February 17, / 105

79 Question (2) Which of the following expressions is equivalent to ( ) log 2 x y (a) log 2 (x 3 ) 1 2 log 2 (y 2 1) (b) 3 2 log 2 (x(y 2 1)) (c) 3 log 2 (x) log 2 (y 2 1) () 3 log 2 (x) log 2 (y 2 ) 1 2 log 2 (1) February 17, / 105

80 Properties of Logarithms Aitional properties that are useful. f (x) = log a (x), has omain (0, ) an range (, ). For a > 1, * lim log x 0 + a (x) = an lim log x a (x) = For 0 < a < 1, lim log x 0 + a (x) = an lim log x a (x) = In avance mathematics (an in light of the change of base formula), we usually restrict our attention to the natural log. February 17, / 105

81 Graphs of Logarithms:Logarithms are continuous on (0, ). Figure: Plots of functions of the type f (x) = log a (x). The value of a > 1 on the left, an 0 < a < 1 on the right. February 17, / 105

82 Examples Evaluate each limit. (a) lim ln(sin(x)) x 0 + (b) lim ln(tan(x)) x π 2 February 17, / 105

83 Questions ( ) 1 (1) Evaluate the limit lim ln x x 2 (a) (b) 0 (c) () The limit oesn t exist. February 17, / 105

84 Logarithms are Differentiable on Their Domain Figure: Recall f (x) = a x is ifferentiable on (, ). The graph of log a (x) is a reflection of the graph of a x in the line y = x. So f (x) = log a (x) is ifferentiable on (0, ). February 17, / 105

85 The Derivative of y = log a (x) To fin a erivative rule for y = log a (x), we use the chain rule. Let y = log a (x), then x = a y. February 17, / 105

86 x log a (x) = 1 x ln(a) Examples: Evaluate each erivative. (a) x log 3 (x) (b) θ log 1 (θ) 2 February 17, / 105

87 Question True or False The erivative of the natural log x ln(x) = 1 x February 17, / 105

88 The function ln x Show that if x < 0, then x ln( x) = 1 x February 17, / 105

89 The Derivative of ln x x ln x = 1 x February 17, / 105

90 Differentiating Functions Involving Logs We can combine our new rule with our existing erivative rules. Chain Rule: Let u be a ifferentiable function. Then x log a (u) = 1 u ln(a) u x = u (x) u(x) ln(a). In particular x ln(u) = 1 u u x = u (x) u(x). February 17, / 105

91 Examples Evaluate each erivative. (a) x ln sec x (b) t t2 log 3 (4t) February 17, / 105

92 Question Fin y if y = x (ln x) 2. (a) y = 2 ln x x (b) y = 2 ln x + 2 (c) y = (ln x) ln x () y = ln(x 2 ) + 2 February 17, / 105

93 Example Determine y x if x ln y + y ln x = 10. February 17, / 105

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95 Using Properties of Logs Properties of logarithms can be use to simplify expressions characterize by proucts, quotients an powers. Illustrative Example: Evaluate ( x 2 ) x ln cos(2x) 3 x 2 + x February 17, / 105

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97 Question Evaluate the erivative. Use properties of logs to simplify the process. ( x 3 ) x ln + 4 x tan x (a) 3x 2 x x 2x +sec2 tan x (b) 3x 2 x x 2x +sec2 tan x (c) 3x 2 x x 2x sec2 tan x February 17, / 105

98 Logarithmic Differentiation We can use properties of logarithms to simplify the process of taking erivatives of expressions that are complicate by proucts quotients an powers. Illustrative Example: Evaluate x ( x 2 ) x + 1 cos 4 (3x) February 17, / 105

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101 Logarithmic Differentiation If the ifferentiable function y = f (x) consists of complicate proucts, quotients, an powers: (i) Take the logarithm of both sies, i.e. ln(y) = ln(f (x)). Then use properties of logs to express ln(f (x)) as a sum/ifference of simpler terms. (ii) Take the erivative of each sie, an use the fact that y x x ln(y) = y. (iii) Solve for y x (i.e. multiply through by y), an replace y with f (x) to express the erivative explicitly as a function of x. February 17, / 105

102 Example Fin y x. y = x 3 (4x 1) 5 4 x + 5 February 17, / 105

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105 Question Fin y x. y = 3 x 2 sin(x) (a) (b) (c) y x = y x = y x = [ ] ( ) 3 2 x 2 sin(x) x + cot x [ ] ( 3 2 x 2 sin(x) 3x + 1 ) 3 cot x [ ] ( 3 1 x 2 sin(x) x sin x ) February 17, / 105