Chapter 3: Transcendental Functions

Transcription

1 Chapter 3: Transcendental Functions Spring 2018 Department of Mathematics Hong Kong Baptist University 1 / 32

2 Except for the power functions, the other basic elementary functions are also called the transcendental functions. Transcendental functions cannot be constructed from real numbers and a single real variable x by using finitely many arithmetic operations (addition, subtraction, multiplication, division, and fractional powers). We have studied the trigonometric functions in Section 2.5. This chapter is devoted to the study of other transcendental functions, including (i) exponential functions, (ii) logarithmic functions, (iii) inverse trigonometric functions (not required). 2 / 32

3 Recall that the elementary functions are built from the basic elementary functions and constants through composition (f g) and combinations using operations (+,, and /). By the rules of differentiation, we conclude that the derivative of any elementary function can be expressed in terms of the basic elementary functions and their derivatives. In this chapter, we will study the differentiation rules for (i) exponential functions and (ii) logarithmic functions, respectively. 3 / 32

4 3.2 Exponential and Logarithmic Functions An exponential function is a function of the form f (x) = a x, where the base a is a positive constant and the exponent x is the variable. Lemma Let a > 0 and b > 0. Then for any pair of real numbers x and y, we have (i) a 0 = 1 (iii) a x = 1 a x (v) (a x ) y = a xy (ii) a x+y = a x a y (iv) a x y = ax a y (vi) (ab) x = a x b x 4 / 32

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6 If a = 1, then a x = 1 x = 1 for every x, i.e., it is a constant function. If a > 1, then a x is an increasing function of x with lim x ax = 0 and lim x ax =. If 0 < a < 1, then a x is a decreasing function of x with lim x ax = and lim x ax = 0. If a > 0 and a 1, then the function f (x) = a x has domain (, ) and range (0, ). In particular, we have a x > 0 for all x. 6 / 32

7 Definition If a > 0 and a 1, then function log a (x), called the logarithm of x to the base a, is the inverse of the one-to-one function a x : y = log a (x) x = a y. log a (x) has domain (0, ) and range (, ). Remark: Since a x and log a (x) are inverse functions, the following cancellation identities hold: log a (a x ) = x for all real x, a log a (x) = x for all x > 0. In particular, log a (a) = 1 when x = 1 and log a (1) = 0 when x = 0. 7 / 32

8 Laws of Logarithms Lemma If x > 0, y > 0, a > 0, b > 0, a 1, and b 1, then (i) log a (1) = 0 (ii) log a (xy) = log a (x) + log a (y) ( ) ( ) 1 x (iii) log a = log x a (x) (iv) log a = log y a (x) log a (y) (v) log a (x y ) = y log a (x) (vi) log a (x) = log b(x) log b (a) 8 / 32

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10 If a > 1, then log a (x) is an increasing function of x with lim log a(x) = and lim log x 0+ x a(x) =. If 0 < a < 1, then log a (x) is a decreasing function of x with lim log a(x) = and lim log x 0+ x a(x) =. 10 / 32

11 Example 1: (a) Simplify 3 log 9 (4) ; (b) Solve the equation 3 x 1 = 2 x. Solution: 11 / 32

12 Example 1: (a) Simplify 3 log 9 (4) ; (b) Solve the equation 3 x 1 = 2 x. Solution: (a) Note that log 9 (4) = log 3(4) log 3 (9) = 1 2 log 3(4). So 3 log 9 (4) = 3 log 3 (4) 1 2 = 4 1/2 = 2. (b) Let a = 2 and take logarithms of both sides, This leads to where log 2 (3) (x 1) log 2 (3) = x. x = log 2(3) log 2 (3) , 12 / 32

13 3.3 The Natural Logarithm and Exponential This section considers the base a = e, where the e constant is called the Euler s number. Specifically, we write log e (x) = ln(x) and e x = exp(x). The e constant was defined by Bernoulli, ( e = lim ) n = lim n n (1 + r 0 r)1/r. The number e is irrational (in fact, transcendental) with value e In 4.10 we will also learn that e = ! + 1 2! + 1 3! + 1 4! + 13 / 32

14 Theorem (1) For any x > 0, d dx ln(x) = 1 x. (2) For any x (, ), d dx ex = e x. 14 / 32

15 Thus, we have d dx ln(x) = 1 x, as required. 15 / 32 Proof: (1) By definition, we have d ln(x) = lim dx h 0 1 = lim h 0 h ln ln(x + h) ln(x) ( h 1 + h ) x = 1 x lim ln(1 + r), r 0 }{{ r } =:L where we let h = rx. But taking the natural logarithm of the definition of e gives ( ) 1 = ln(e) = ln (1 + r)1/r lim r 0 1 = lim ln(1 + r) = L. r 0 r

16 (2) The derivative of y = e x can be calculated by implicit differentiation. Note that y = e x = x = ln(y). By the Chain rule and take derivatives of both sides on x, 1 = 1 y dy dx. This leads to dy dx = y = ex. 16 / 32

17 Example 2: Find the derivatives of (a) ln(cos 2 (x)), and (b) ln(x + x 2 + 1). Solution: 17 / 32

18 Example 2: Find the derivatives of (a) ln(cos 2 (x)), and (b) ln(x + x 2 + 1). Solution: (a) By the Chain rule, d dx ln(cos2 (x)) = (b) By the Chain rule, 1 cos 2 (x) d dx cos2 (x) = 2 tan(x). d dx ln(x + x 2 + 1) = = = 1 x + x d dx (x + x 2 + 1) ( ) 1 x + 2x 1 + x x x / 32

19 Example 3: Find the derivatives of (a) e x2 3x, and (b) 1 + e 2x. Solution: 19 / 32

20 Example 3: Find the derivatives of (a) e x2 3x, and (b) 1 + e 2x. Solution: (a) By the Chain rule, d 3x dx ex2 = e x2 3x d dx (x 2 3x) = (2x 3)e x2 3x. (b) By the Chain rule, d 1 + e dx 2x = e 2x d dx (1 + e2x ) = e 2x 1 + e 2x. 20 / 32

21 General Exponentials and Logarithms Theorem (1) For any x (, ), d dx ax = d dx ex ln(a) = e x ln(a) ln(a) = a x ln(a). (2) For any x > 0, d dx log a(x) = 1 x ln(a). Remark: The derivative of log a (x) can be derived (i) by implicit differentiation on x = a y, or (ii) by noting that log a (x) = ln(x) ln(a). 21 / 32

22 Functions y = e x y = a x y = ln(x) Derivatives y = e x y = a x ln(a) y = 1 x y = log a (x) y = 1 x ln(a) 22 / 32

23 Suppose we want to differentiate a function of the form y = (f (x)) g(x) (for f (x) > 0). For instance, when the function is y = x x. Since the variable appears in both the base and the exponent, neither the general power rule nor the exponential rule can be directly applied. One method for finding the derivative of such a function is to express it in the form g(x) ln(f (x)) y = e and then differentiate, using the Product rule to handle the exponent. 23 / 32

24 Logarithmic Differentiation Another method to obtain the derivative of such a function is to take the natural logarithms of both sides of the equation y = (f (x)) g(x) and then to differentiate it implicitly: ln(y) = g(x) ln f (x) 1 dy y dx = g (x) ln f (x) + g(x) f (x) f (x) ( dy dx = y g (x) ln f (x) + g(x) ) f (x) f (x) = (f (x)) g(x) ( g (x) ln f (x) + g(x) f (x) f (x) This technique is called logarithmic differentiation. ). 24 / 32

25 Example 4: Find the derivative of y = x x. Solution 1: Note that y = e x ln(x). By the Chain rule and the Product rule, dy dx = d dx ex ln(x) = e x ln(x) d dx (x ln(x)) = x x (ln(x) + 1). Solution 2: Let f (x) = x and g(x) = x. By logarithmic differentiation, ( = (f (x)) g(x) g (x) ln f (x) + g(x) dy dx = x x (ln(x) + 1). ) f (x) f (x) 25 / 32

26 3.5 Inverse Trigonometric Functions The inverse trigonometric functions are inverse functions of sin, cos, tan, etc. For example, the arcsine function satisfies y = arcsin(x) = x = sin(y). Other inverse trigonometric functions are define similarly: y = arccos(x) = x = cos(y), y = arctan(x) = x = tan(y). 26 / 32

27 y = arcsin(x) = x = sin(y). Since 1 sin(y) 1 for all y, the arcsine function is only defined for x [ 1, 1]. Since the equation x = sin(y) has many solutions for a given x, we restrict the range of arcsine to the interval [ π/2, π/2]. 27 / 32

28 Lemma (Cancellation Identity) For x [ π/2, π/2] and y [ 1, 1], we have arcsin(sin(x)) = x, sin(arcsin(y)) = y. Examples: (a) arcsin( 1 2 ) = π/6; (b) arcsin(2) is not defined; (c) sin(arcsin( 1 2 )) = 1 2 ; (d) arcsin(sin( 7π 6 )) = π / 32

29 Theorem (Derivative of arcsine) For 1 < x < 1, we have d dx arcsin(x) = 1. 1 x 2 Proof: Let y = arcsin(x) = sin(y) = x, π/2 < y < π/2. Then implicit differentiation gives cos(y) dy dx = 1 = dy dx = 1 cos(y). But cos(y) is positive for π/2 < y < π/2, so that cos 2 (y) = 1 sin 2 (y) = 1 x 2 = cos(y) = 1 x 2. Thus, dy dx = 1 1 x / 32

30 Example 5: Find the derivative of arcsin( x a ) and hence evaluate dx, where a > 0. a 2 x 2 Solution: By the chain rule, d ( x ) dx arcsin = a (x/a) 2 a = 1 a 2 x. 2 Hence, dx ( x ) a 2 x = arcsin + C. 2 a 30 / 32

31 The arccosine function is defined for x [ 1, 1], with the range [0, π]. Theorem (Derivatives of arccos) For 1 < x < 1, d dx arccos(x) = 1. 1 x 2 31 / 32

32 The arctangent function is defined for all x R, with range [ π/2, π/2]. Theorem (Derivatives of arctan) d dx arctan(x) = x / 32