MATH 3333{INTERMEDIATE ANALYSIS{BLECHER NOTES 55 Abbreviated notes version for Test 3: just the definitions, theorem statements, proofs on the list. I may possibly have made a mistake, so check it. Also, this is not intended as a REPLACEMENT for your classnotes; the classnotes have lots of other things that you may need for your understanding, like worked examples. 6. The derivative 6.1. Differentiation rules. Definition: Let f :(a; b) ! R and c 2 (a; b). If the limit lim f(x)−f(c) exists and is finite, then we say that f is differentiable at x!c x−c 0 df c, and we write this limit as f (c) or dx (c). This is the derivative of f at c, and also obviously equals lim f(c+h)−f(c) by setting x = c + h or h = x − c. If f is h!0 h differentiable at every point in (a; b), then we say that f is differentiable on (a; b). Theorem 6.1. If f :(a; b) ! R is differentiable at at a point c 2 (a; b), then f is continuous at c. Proof. If f is differentiable at c then f(x) − f(c) f(x) = (x − c) + f(c) ! f 0(c)0 + f(c) = f(c); x − c as x ! c. So f is continuous at c. Theorem 6.2. (Calculus I differentiation laws) If f; g :(a; b) ! R is differentiable at a point c 2 (a; b), then (1) f(x) + g(x) is differentiable at c and (f + g)0(c) = f 0(c) + g0(c). (2) f(x) − g(x) is differentiable at c and (f − g)0(c) = f 0(c) − g0(c). (3) Kf(x) is differentiable at c if K is a constant, and (Kf)0(c) = Kf 0(c). (4) (Product rule) f(x)g(x) is differentiable at c, and (fg)0(c) = f 0(c)g(c) + f(c)g0(c). f(x) f 0 (5) (Quotient rule) g(x) is differentiable at c if g(c) 6= 0, and ( g ) (c) = f 0(c)g(c)−f(c)g0(c) g(c)2 . Proof. In the last step of many of the proofs below we will be silently using the definition of the derivative, and the `limit laws' from Theorem 5.2. (1) As x ! c we have f(x) + g(x) − (f(c) + g(c) f(x) − f(c) g(x) − g(c) = + ! f 0(c) + g0(c): x − c x − c x − c (4) We have f(x)g(x) − f(c)g(c) f(x)g(x) − f(x)g(c) f(x)g(c) − f(c)g(c) g(x) − g(c) f(x) − f(c) = + = f(x) +g(c) : x − c x − c x − c x − c x − c By Theorem 6.1, f is continuous at c, that is, limx!c f(x) = f(c). So f(x)g(x) − f(c)g(c) g(x) − g(c) f(x) − f(c) lim = lim f(x) +g(c) = f(c)g0(c)+g(c)f 0(c): x!c x − c x!c x − c x − c 56 MATH 3333{INTERMEDIATE ANALYSIS{BLECHER NOTES (3) Set g(x) = K in (4), to get (Kf)0(c) = Kf 0(c) + 0f(c) = Kf 0(c). (2) By (1) and (3), (f + (−g))0 = f 0 + (−g)0 = f 0 − g0. (5) Since g(c) 6= 0, so that jg(c)j > 0, by Proposition 5.7 jg(x)j, and hence also g(x), is nonzero on a neighborhood of c. So division by g(x) in what follows is justified. We have f(x) f(c) − f(x)g(c) − g(x)f(c) f(x)g(c) − f(c)g(c) + f(c)g(c) − f(c)g(x) g(x) g(c) = = : x − c g(x)g(c)(x − c) g(x)g(c)(x − c) This equals f(x)−f(c) g(x)−g(c) g(c) − f(c) g(c)f 0(c) − f(c)g0(c) x−c x−c ! g(x)g(c) (g(c))2 as x ! c, since by Theorem 6.1, limx!c g(x) = g(c). Theorem 6.3. (Calculus I chain rule) If f :(a; b) ! R is differentiable at a point c 2 (a; b), and if g : I ! R is differentiable at f(c), where I is an open interval containing f((a; b)), then the composition g ◦ f is differentiable at c and (g ◦ f)0(c) = g0(f(c))f 0(c). As in Calculus, a point c is called a local minimum (resp. local maximum) point for a function f if 9 a neighborhood U of c s.t. f(c) ≤ f(x) (resp. f(c) ≥ f(x)) 8x 2 V . In this case we say that f(c) is a local minimum (resp. maximum) value of f. As in Calculus I, the word `extreme' means either `minimum' or `maximum'. So we have local extreme points and local extreme values, just as in Calculus I. Recall from Calculus that a critical point is a point c s.t. f 0(c) does not exist or f 0(c) = 0. Theorem 6.4. (First Derivative Test) If f :(a; b) ! R is differentiable at a local extremum point c 2 (a; b), then f 0(c) = 0 (and so c is a critical point). Proof. Suppose that c is a local maximum point of f, so f(x) ≤ f(c) for all x in a f(x) − f(c) neighborhood of c. Then lim ≤ 0 (by the bullet just before Propo- x!c+ x − c | {z } ≤0 f(x) − f(c) sition 5.4 with g = 0). Similarly, lim ≥ 0. Since f is differentiable x!c− x − c | {z } ≥0 f 0(c) = lim f(x)−f(c) exists, and so these two one-sided limits must be equal by x!c x−c Proposition 5.4, and hence must be 0. That is, f 0(c) = lim f(x)−f(c) = 0. The x!c x−c local minimum case is similar. 6.2. The mean value theorem. Lemma 6.5. (Rolle's theorem) If f :[a; b] ! R is continuous, and f is differen- tiable on (a; b), and f(a) = f(b) = 0, then there exists c 2 (a; b) with f 0(c) = 0. MATH 3333{INTERMEDIATE ANALYSIS{BLECHER NOTES 57 Proof. To prove this, note that it is clearly true if f is constant. By the MAX- MIN theorem f has a maximum and a minimum value on [a; b]. At least one of these extreme values must be nonzero if f is not constant, and hence this value is achieved at a point c which is not a or b. By the first derivative test (Theorem 6.4), 0 f (c) = 0. Theorem 6.6. (The mean value theorem) If f :[a; b] ! R is continuous, and f is differentiable on (a; b), then there exists c 2 (a; b) with f 0(c) = (f(b)−f(a))=(b−a). Equivalently, f(b) − f(a) = f 0(c)(b − a). f(b)−f(a) 0 f(b)−f(a) Proof. Let g(x) = b−a (x − a) + f(a). Clearly g = b−a . Then h = f − g is continuous on [a; b] and differentiable on (a; b). Also it is easy to check that h(a) = h(b) = 0, so by Rolle's theorem 9c 2 (a; b) s.t. h0(c) = f 0(c) − g0(c) = 0. 0 0 f(b)−f(a) Hence f (c) = g (c) = b−a . Corollary 6.7. If f :(a; b) ! R has f 0(x) = 0 for all x 2 (a; b), then f is constant on (a; b). Proof. If a < x < y < b then by the MVT there exists c such that f(y) − f(x) = 0 f (c)(y − x) = 0. Thus f(x) = f(y). That is, f is constant. Corollary 6.8. If f :(a; b) ! R has f 0(x) > 0 for all x 2 (a; b) then f(x) is strictly increasing on (a; b). Similarly, if f 0(x) ≥ 0 (resp. f 0(x) < 0; f 0(x) ≤ 0) for all x 2 (a; b) then f(x) is increasing (resp. strictly decreasing, decreasing) on (a; b). Proof. We just prove one, the others are similar. Suppose that f 0(x) > 0 for all x 2 (a; b). If a < x < y < b then by the MVT there exists c such that f(y) − f(x) = f 0(c)(y − x) > 0, since f 0(c) > 0. Thus f(x) < f(y). So f is strictly increasing on (a; b). Corollary 6.9. If f 0(x) = g0(x) for all x 2 (a; b) then there exists a constant C such that f(x) = g(x) + C for all x 2 (a; b). 6.3. Taylors theorem and the second derivative test. Here is a version of the Calculus II Taylor's theorem: Theorem 6.10. Suppose that n 2 N (or n = 0). If f :(a; b) ! R is n + 1 times differentiable, and the first n of these derivatives are continuous on (a; b), and if x0 2 (a; b), then for every x 2 (a; b) with x 6= x0 there is a number c between x0 and x such that n (k) (n+1) X f (x0) f (c) f(x) = (x − x )k + (x − x )n+1: k! 0 (n + 1)! 0 k=0 58 MATH 3333{INTERMEDIATE ANALYSIS{BLECHER NOTES Remark.