6.1. The Derivative Definition (6.1.1). Let I R be an interval, f : I! R, and c 2 R. We say L is the derivative of f at c if

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1 CHAPTER 6 Di erentiation 6.1. The Derivative Definition (6.1.1). Let I R be an interval, f : I! R, and 2 R. We say L is the derivative of f at if f(x) > 0 9 > 0 3 x 2 I and 0 < < =) L <. In this ase, we say f is di erentiable at, and write f 0 () for L. Note. (1) f 0 f(x) () = lim provided the limit exists. (2) If is an endpoint of I, use lim or lim as appropriate. + (3) We obtain a funtion f 0 where D f 0 D f. (4) For h =, f 0 f( + h) () = lim provided the limit exists. h!0 h Example. < x 2 sin 1 (1) f(x) = x, x 6= 0 : 0, x = 0 sine Thus f 0 (0) = 0. f 0 (0) = lim x!0 f(x) f(0) x 0 = lim x!0 x 2 sin 1 x x x sin x 0 = x sin 1 x 104 = lim x sin 1 x!0 x apple x! 0. = 0

2 6.1. THE DERIVATIVE 105 Theorem (6.1.2). If f : I! R has a derivative at 2 I, then f is ontinuous at. Proof. lim [f(x) apple f(x) ] = lim x lim f(x) x () = lim () = f 0 () 0 = 0 =) lim f(x) = =) f is ontinuous at.

3 DIFFERENTIATION Example. (2) The Ruler (Thomae s) funtion < 0, if x 62 Q f(x) = 1 : n, if x 2 Q with x = m n, n 2 N, m. in lowest terms n Sine f is disontinuous at eah rational, f is not di erentiable at any rational. (graph is on Page 127). Suppose is irrational. Then = d.b 1 b 2 b 3..., a nonrepeating deimal with d 2 Z, b n 2 Z, 0 apple b n apple 9. Let Then (x n )! and f(x n ) x n = {z } n x n = d.b 1 b 2... b n. f(x n ) b n+1 b n n b n+1 b n+2... = 1.b n+1 b n+2... > 1. Now let (y n ) be a sequene of irrationals 3 (y n )!. then f(x) Thus lim x f(y n ) y n = 0 0 y n = 0. DNE anywhere, so f 0 () DNE anywhere. Note. Example (2) implies that Theorem has no onverse. 1X 1 In fat, f(x) = 2 n os(3n x) is ontinuous everywhere, but di erentiable n=0 nowhere.

4 6.1. THE DERIVATIVE 107 Theorem (6.1.3). Let I R be an interval, 2 I, and f : I! R and g : I! R be di erentiable at. Then (a) If 2 R, f is di erentiable at and ( f) 0 () = f 0 (). (b) f + g is di erentiable at and (f + g) 0 () = f 0 () + g 0 (). () fg is di erentiable at and (fg) 0 () = f 0 ()g() + g 0 (). (d) If g() 6= 0, f f 0() g is di erentiable at and f 0 ()g() g 0 () =. g [g()] 2 Proof. (d) Sine g() 6= 0, 9 an interval J I 3 g(x) 6= 0 x 2 J. Then f g 0() = lim lim lim f g (x) f g f(x) g(x) g() () = lim = f(x)g() g(x) lim = g(x)g()() f(x)g() g() + g() g(x) = g(x)g()() apple f(x) g(x) g() g() = f 0 ()g() g 0 (). [g()] 2 1 g(x)g()

5 10 6. DIFFERENTIATION Theorem (6.1.6 Chain Rule). Let I, J be intervals in R, g : I! R and f : J! R 3 f(j) I, and 2 J. If f is di erentiable at and g is di erentiable at, then g f is di erentiable at and (g f) 0 () = g 0 f 0 (). Note. g f(x) g = g f(x) g f(x) seems to suggest the orret result. But what if f(x) = near? Proof. Let d = and G : I! R be defined by < g(y) g(d), y 2 I, y 6= d G(y) = y d. : g 0 (d), y = d Sine g is di erentiable at d, lim y!d G(y) = g 0 (d) = G(d), f(x) so G is ontinuous at d. Sine f is ontinuous at and f(j) 2 I, G f is ontinuous at =) lim (G f)(x) = (G f)() = g0. Now g(y) g(d) = G(y)(y d) y 2 I. Thus, for x 2 J and y = f(x) 2 I, Then (g f)(x) (g f)() = (G f)(x)( f(x). (g f) 0 (g f)(x) (g f)() () = lim = apple f(x) lim (G f)(x) lim (G f(x) f)(x) lim x = = g 0 f 0 ().

6 Example THE DERIVATIVE 109 (3) Suppose f : I! R is di erentiable on I and g(y) = y n y 2 R, n 2 N. Then g 0 (y) = ny n or 1, so (g f) 0 (x) = g 0 f(x) f 0 (x) for x 2 I (f n ) 0 (x) = n f(x) n 1 f 0 (x) for x 2 I. (4) f(x) = ( x 2 sin 1 x, x 6= 0. 0, x = 0 For x 6= 0, f 0 (x) = 2x sin 1 x + x2 os 1 x 1 = 2x sin 1 x 2 x From Example (1), f 0 (0) = 0, but lim f 0 (x) DNE, x!0 so f 0 is not ontinuous at 0 =) f 00 (0) DNE. Note that f 00 (x) exists for x 6= 0. os 1 x.

7 DIFFERENTIATION (5) f(x) = x + x + 1. x < 1 =) f(x) = x (x + 1) = 2x 1 =) f 0 (x) = 2. 1 < x < 0 =) f(x) = x + (x + 1) = 1 =) f 0 (x) = 0. x > 0 =) f(x) = x + (x + 1) = 2x + 1 =) f 0 (x) = f(x) f( 1) >< x + 1 = 0, x > 1 x = 1 =) = =) f 0 ( 1)DNE x ( 1) >: 2x 1 1 = 2, x < 1 x + 1 2x x = 0 =) f(x) f(0) >< = 2, x > 0 x = =) f 0 (0)DNE x 0 >: 1 1 = 0, x < 0 x Homework Pages # 1d, 2, 4, b

8 6.2. THE MEAN VALUE THEOREM The Mean Value Theorem Definition. f : I! R has a relative maximum [resp., minimum] at 2 I if 9 a neighborhood V () 3 f(x) apple [resp., f(x) ] x 2 V () \ I. A relative extremum at is a relative maximum or minimum at. Theorem (6.2.1 Interior Extremum Theorem). Let be an interior point of the interval I 3 f : I! R has a relative extremum at. Then, if f 0 () exists, f 0 () = 0. Proof. We assume f has a relative minimum at. The other ase is similar. If f 0 () > 0, f(x) 9 > < < =) > 0 by Theorem But f(x) 0 for x near, so x < provides a ontradition. If f 0 () < 0, f(x) 9 > < < =) < 0 by Theorem Again, sine f(x) 0 for x near, x > provides a ontradition. Thus f 0 () = 0.

9 DIFFERENTIATION Theorem (6.2.3 Rolle s Theorem). Suppose f is ontinuous on [a, b], di erentiable on (a, b), and f(a) = f(b). Then 9 2 (a, b) 3 f 0 () = 0. Proof. By the Maximum-Minimum Theorem 9 x 0, y 0 2 [a, b] 3 f(x 0 ) apple f(x) apple f(y 0 ) If x 0 and y 0 are endpoints of [a, b], then f is onstant and f 0 (x) = 0 x 2 (a, b). x 2 [a, b]. Otherwise, f has a maximum or minimum at some 2 (a, b), so f 0 () = 0 by the Interior Extremum Theorem.

10 6.2. THE MEAN VALUE THEOREM 113 Theorem (6.2.4 Mean Value Theorem (MVT)). Suppose f is ontinuous on [a, b], di erentiable on (a, b). Then 9 2 (a, b) 3 f(b) f(a) = f 0 ()(b a) or f 0 () = f(b) f(a) b a for a 6= b. Proof. Consider : [a, b]! R defined by f(b) f(a) (x) = f(x) f(a) (x a). b a is ontinuous on [a, b], di erentiable on (a, b), and (a) = (b) = 0. By Rolle s theorem 9 2 (a, b) 3 0 () = 0. Sine 0 (x) = f 0 f(b) f(a) (x), b a f 0 () = f(b) f(a). b a Note. The MVT allows us to get information about f from f 0.

11 DIFFERENTIATION Theorem (6.2.5). Suppose f is ontinuous on [a, b], di erentiable on (a, b), and f 0 (x) = 0 x 2 (a, b). Then f is onstant on [a, b]. Proof. To show f(x) = f(a) x 2 [a, b]. For x > a, apply the MVT to f on [a, x]. Then Sine f 0 () = 0, f(x) = f(a). 9 2 (a, x) 3 f(x) f(a) = f 0 ()(x a). Corollary (6.2.6). Suppose f and g are ontinuous on [a, b], di erentiable on (a, b), and f 0 (x) = g 0 (x) on (a, b). Then 9 C 2 R 3 f(x) = g(x) + C on [a, b]. Proof. Apply Theorem to f g. Then C = f(a) g(a). Theorem (6.2.7). Let f : I! R be di erentiable on an interval I. Then (a) f is inreasing (i.e., nondereasing) on I () f 0 (x) 0 on I. (b) f is dereasing (i.e., noninreasing) on I () f 0 (x) apple 0 on I. Proof. (a) ((=) Suppose x 1, x 2 2 I 3 x 1 < x 2. Applying the MVT to f on [x 1, x 2 ], 9 2 (x 1, x 2 ) 3 f(x 2 ) f(x 1 ) = f 0 ()(x 2 x 1 ). Sine f 0 () 0 and x 2 x 1 > 0, f(x 2 ) f(x 1 ) 0 =) f(x 1 ) apple f(x 2 ). (=)) Let 2 I. Suppose x > or x < for x 2 I. Then (b) Similar. f(x) x 0 =) f 0 f(x) () = lim x 0.

12 Example. g(x) = < x + 2x 2 sin 1 x, x 6= 0. : 0, x = THE MEAN VALUE THEOREM 115 It an be shown that g 0 (0) = 1, but g is not inreasing in any neighborhood of 0. [Thus the notion of inreasing and dereasing only apply to intervals.] For x 6= 0, g 0 (x) = 1 + 4x sin 1 x 2 os 1 x. g 0 (0) = lim x!0 g(x) g(0) x 0 1 g 0 2n = n = lim x!0 x + 2x 2 sin 1 x x 0 g = (4n + 1) (4n + 1) = lim 1+2x sin 1 x!0 x sin 2n 2 os 2n = = 1. sin (4n + 1) 2 = 1+0 = 1. (4n + 1) 2 os = > 0. (4n + 1) Thus in any neighborhood of 0, we have both g 0 (x) 0 and g 0 (x) apple 0. Homework Pages # 6, 7