4. We accept without proofs that the following functions are differentiable: (e x ) = e x, sin x = cos x, cos x = sin x, log (x) = 1 sin x

Transcription

1 4 We accept without proofs that the following functions are differentiable: (e x ) = e x, sin x = cos x, cos x = sin x, log (x) = 1 sin x x, x > 0 Since tan x = cos x, from the quotient rule, tan x = sin x cos x sin x cos x = cos2 x+sin 2 x = 1 cos 2 x cos 2 x cos 2 x sec2 x if cos x 0 Similarly, cot x = 1 sin 2 x = csc2 x Theorem 413 [Chain Rule] If f is differentiable at a and g is differentiable at f(a), then the composite function g f is differentiable at a and (g f) (a) = g (f(a)) f (a) Proof Define h on dom(g) such that h(f(a)) = g (f(a)), and h(y) = g(y) g(f(a)), y dom(g), y f(a) y f(a) From the definition of g, h is continuous at f(a) For any x dom(f) with x a, we always have g f(x) g f(a) f(x) f(a) = h(f(x)) x a x a Check: if f(x) = f(a), both sides equal to 0; if f(x) f(a), then h(f(x)) = g(f(x)) g(f(a)) f(x) f(a) Thus, g f(x) g f(a) f(x) f(a) lim = lim h(f(x)) lim = h(f(a)) f (a) = g (f(a)) f (a), x a x a x a x a x a where the second last equality follows from that f is continuous at a and h is continuous at f(a) Thus, g f is differentiable at a, and (g f) (a) = g (f(a)) f (a) Examples 1 Let h(x) = sin(x 3 + 7x) Then h = g f, where f(x) = x 3 + 7x and g(x) = sin x We know that g (x) = cos x and f (x) = 3x From the Chain rule, we have h (x) = g (f(x)) f (x) = cos(x 3 + 7x) (3x 2 + 7) 2 Let h(x) = e cos x+x2 Then h = g f, where f(x) = cos x + x 2 and g(x) = e x We know that g (x) = e x and f (x) = sin x + 2x From the Chain rule, we have h (x) = g (f(x)) f (x) = e cos x+x2 ( sin x + 2x) 3 Let h(x) = log x for x 0 If x > 0, h (x) = 1 x If x < 0, h(x) = log( x), so from the Chain rule, h (x) = 1 x ( 1) = 1 x Thus, h (x) = 1 x for any x 0 4 If f is differentiable, its derivative function f may not be continuous An example is given by f(x) = x 2 sin(1/x) for x 0, and f(0) = 0 Such f is differentiable on R, but f is not continuous at 0 See Exercise 284 Definition 412 Let f be differentiable 66

2 1 If f is continuous, we say that f is continuously differentiable, or f is C 1 2 If f is differentiable, we say that f is twice differentiable, and write f = (f ) 3 If f is twice differentiable, and f is continuous, we say that f is twice continuously differentiable, or f is C 2 4 If f is twice differentiable, and f is is differentiable, we say that f is three times differentiable, and write f = (f ) 5 In general, we write f (0) = f and f (1) = f Whenever f (n) exists and is differentiable, we define f (n+1) = (f (n) ) If f (n) exists, we say that f is n times differentiable; if f (n) is continuous, we say that f is n times continuously differentiable, or f is C n 6 If f (n) exists for any n N, we say that f is C, or f is smooth Examples 1 f(x) = e x is smooth because f (n) (x) = e x for all n N 2 cos x and sin x are smooth because sin x = cos x and cos x = sin x 3 Every polynomial is smooth because if the degree of P is d, then P (d+1) is constant zero, and so do P (n) for any n d + 1 Homework 284, 286, 289 Let f(x) = e ax cos(bx) and g(x) = e ax sin(bx), where a, b R (a) Compute f and g (b) Use (a) to compute f and f 42 Mean Value Theorem Theorem 421 Let f be defined on (a, b) which contains x 0 Suppose f assumes its maximum or minimum at x 0, and f is differentiable at x 0 Then f (x 0 ) = 0 Proof First, suppose that f assumes its maximum at x 0 This means that f(x 0 ) f(x) for any x (a, b) To prove that f (x 0 ) = 0, we show that it is not possible that f (x 0 ) > 0 or f (x 0 ) < 0 Suppose f (x 0 ) > 0 Let ε = f (x 0 )/2 From the definition of f (x 0 ), there is δ > 0 such that if x (x 0 δ, x 0 ) (x 0, x 0 + δ), f(x) f(x 0) f (x 0 ) < ε, x x 0 which implies that f(x) f(x 0 ) > f (x 0 ) ε = f (x 0 ) > 0 x x

3 So if x (x 0, x 0 + δ), then f(x) f(x 0 ) > 0, which contradicts that f assumes its maximum at x 0 Thus, f (x 0 ) > 0 is impossible Now suppose f (x 0 ) < 0 Let ε = f (x 0 )/2 > 0 Then there is δ > 0 such that if x (x 0 δ, x 0 ) (x 0, x 0 + δ), which implies that f(x) f(x 0) f (x 0 ) < ε, x x 0 f(x) f(x 0 ) < f (x 0 ) + ε = f (x 0 ) < 0 x x 0 2 So if x (x 0 δ, x 0 ), then f(x) f(x 0 ) > 0, which again is a contradiction This means that f (x 0 ) < 0 is also not possible Thus, f (x 0 ) = 0 Finally, if f assumes its minimum at x 0, then f assumes its maximum at x 0 Moreover, f is also differentiable at x 0 From above, ( f) (x 0 ) = 0 Thus, f (x 0 ) = ( f) (x 0 ) = 0 Remark The condition that f is differentiable at x 0 can not be omitted For example, f(x) = x attains its minimum at 0, but f is not differentiable at 0 Theorem 422 [Rolle s Theorem] Let f be continuous on [a, b] and differentiable on (a, b) Suppose f(a) = f(b) Then there is c (a, b) such that f (c) = 0 Proof Since f is continuous on [a, b], from Extreme Value Theorem, there exist x 0, y 0 [a, b] such that f(x 0 ) f(x) f(y 0 ) for all x [a, b] If x 0 (a, b), then we let c = x 0 From the previous theorem, f (x 0 ) = 0 If y 0 (a, b), then we let c = y 0 From the previous theorem, f (c) = 0 The remaining case is x 0, y 0 {a, b} Since f(a) = f(b), we have f(x 0 ) = f(y 0 ) Since f(x 0 ) f(x) f(y 0 ) for all x [a, b], f is constant on [a, b], so f = 0 on (a, b) Then we may let c = (a + b)/2 Theorem 423 [Mean Value Theorem] Let f be continuous on [a, b] and differentiable on (a, b) Then there is c (a, b) such that f (c) = f(b) f(a) b a Proof Let k = f(b) f(a) b a R and g(x) = f(x) kx for x [a, b] Then g is continuous on [a, b] and differentiable on (a, b) And f (x) = g (x) + k for x (a, b) We compute g(b) g(a) = f(b) f(a) k(b a) = f(b) f(a) (f(b) f(a)) = 0 So g(a) = g(b) From Rolle s Theorem, there is c (a, b) such that g (c) = 0 Thus, f (c) = g (c) + k = k = f(b) f(a) b a 68

4 Remark The value of f(b) f(a) b a is the slope of the line connecting (a, f(a)) and (b, f(b)), while f (c) is the slope of the tangent line to the curve y = f(x) at (c, f(c)) There may exist more than one c (a, b) such that f (c) = f(b) f(a) b a Remark We will see some applications of MVT In those applications, we will first know some properties of the derivative function f, and then use MVT to derive some properties of f itself Corollary 421 Let f be differentiable on (a, b) such that f (x) = 0 for all x (a, b) Then f is a constant function on (a, b) Proof It suffices to show that for any x 1 < x 2 (a, b), f(x 1 ) = f(x 2 ) From the MVT, there exists c (x 1, x 2 ) such that f(x 2) f(x 1 ) x 2 x 1 = f (c) = 0, which implies that f(x 2 ) f(x 1 ) = 0 Corollary 422 Let F and G be differentiable on (a, b) such that F (x) = G (x) for all x (a, b) Then there is c R such that G(x) = F (x) + c for x (a, b) Proof Let f = G F Then f is differentiable on (a, b) and f (x) = F (x) G (x) = 0 for all x (a, b) From the previous corollary, f is constant, say c R Then G(x) = F (x) + f(x) = F (x) + c, x (a, b) Remark This result is very important in the theory of differential equations Suppose we have a function f defined on (a, b) We want to find a differentiable function F on (a, b) such that F = f Such F is called an antiderivative of f For example, sin x is an antiderivative of cos x on R The above corollary says that if f has one antiderivative F, then we can find all antiderivatives of f, which are F + C, C R For example, every antiderivative of cos x on R has the form of sin x + C for some C R Corollary 423 Let f be differentiable on (a, b) Suppose f is bounded on (a, b) Then f is uniformly continuous on (a, b) Proof We may find M R with M > 0 such that f (x) M for all x (a, b) Let x < y (a, b) From MVT, there is c (x, y) such that f(y) f(x) y x = f (c) Thus, f(y) f(x) = f (c) y x M y x This inequality stays the same if we switch x and y So it also holds if y < x It is also true if x = y since in that case both sides equal to 0 Given ε > 0, let δ = ε M > 0 If x, y (a, b) with x y < δ, then f(y) f(x) M y x < Mδ = ε So f is uniformly continuous on (a, b) Remark The function f that satisfies f(y) f(x) = f (c) y x M y x for all x, y dom(f) is called a Lipschitz function Definition 421 Let f be a function with domain S 1 We say that f is increasing if x 1, x 2 S with x 1 < x 2, f(x 1 ) f(x 2 ) 2 We say that f is decreasing if x 1, x 2 S with x 1 < x 2, f(x 1 ) f(x 2 ) 69

5 3 We say that f is monotonic if it is increasing or decreasing 4 We say that f is strictly increasing if x 1, x 2 S with x 1 < x 2, f(x 1 ) < f(x 2 ) 5 We say that f is strictly decreasing if x 1, x 2 S with x 1 < x 2, f(x 1 ) > f(x 2 ) 6 We say that f is strictly monotonic if it is strictly increasing or strictly decreasing Remark If f is strictly monotonic, then it is injective, ie, one-to-one On the other hand, if f is continuous and injective on an open interval I, then f must be strictly monotonic In fact, if f is not strictly monotonic, then one can find a < b < c I such that either f(a) > f(b) and f(b) < f(c) or f(a) < f(b) and f(b) > f(c) In either case, using IVT, one can find x 1 (a, b) and x 2 (b, c) such that f(x 1 ) = f(x 2 ) Corollary 424 Let f be continuous on [a, b] and differentiable on (a, b) Then 1 f is increasing if f (x) 0 for all x (a, b) 2 f is strictly increasing if f (x) > 0 for all x (a, b) 3 f is decreasing if f (x) 0 for all x (a, b) 4 f is strictly decreasing if f (x) < 0 for all x (a, b) Proof 1 Consider x 1 < x 2 (a, b) From MVT, there is c (x 1, x 2 ) such that f(x 2) f(x 1 ) x 2 x 1 = f (c) 0, which implies that f(x 1 ) f(x 2 ) The remaining cases are similar Corollary 425 Let f be continuous on [a, b] and differentiable on (a, b) 1 If f (x) 0 for all x (a, b), then f attains its maximum on [a, b] at b and its minimum on [a, b] at a 2 If f (x) 0 for all x (a, b), then f attains its maximum on [a, b] at a and its minimum on [a, b] at b Corollary 426 Let f be continuous on [a, b] and differentiable on (a, b) Let c (a, b) 1 If f 0 on (a, c) and f 0 on (c, b), then f assumes its maximum on (a, b) at c 2 If f 0 on (a, c) and f 0 on (c, b), then f assumes its minimum on (a, b) at c Examples 1 f(x) = e x, f (x) = e x > 0 for all x R So f is strictly increasing on R 2 f(x) = 1 x, f (x) = 1 x 2 < 0 for all x R \ {0} So f is strictly decreasing on (, 0) and on (0, ) But it is not strictly decreasing on (, 0) (0, ) because f( 1) = 1 < 1 = f(1) 70

6 3 f(x) = sin x, f (x) = cos(x) For any n Z, cos(x) > 0 on (2nπ π/2, 2nπ + π/2), and cos x < 0 on [2nπ + π/2, 2nπ + 3π/2] Thus, sin x is strictly increasing on (2nπ π/2, 2nπ + π/2), and strictly decreasing on [2nπ + π/2, 2nπ + 3π/2] 4 f(x) = e x x 1 We have f (x) = e x 1 = e x e 0 Since e x is strictly increasing, f (x) > f(0) on (0, ), and f (x) < f(0) on (, 0) Thus, f reaches its minimum at 0, which is f(0) = e = 0 So f(x) 0 for all x R We then conclude that e x x + 1 for all x R Thus, lim x e x 1 = and lim x e = 0 We also have x lim x e x = lim x e x 1 = lim x e = 0 x Homework 291, 292, 297, 299, 2911 Theorem 424 [Intermediate Value Theorem for Derivatives] Let f be a differentiable function on (a, b) Suppose x 1 < x 2 (a, b), and a number c lies between f (x 1 ) and f (x 2 ) Then there exists x 0 (x 1, x 2 ) such that f (x 0 ) = c Proof We may assume that f (x 1 ) < c < f (x 2 ) The case that f (x 1 ) > c > f (x 2 ) is similar Let g(x) = f(x) cx Then g is differentiable function on (a, b) and g (x) = f (x) c Thus, g (x 1 ) < 0 < g (x 2 ) Now g is continuous on [x 1, x 2 ] So there is x 0 [x 1, x 2 ] such that g(x 0 ) g(x) for any x [x 1, x 2 ] If x 0 (x 1, x 2 ), then we must have g (x 0 ) = 0, which implies that f (x 0 ) = c We now exclude the possibility that x 0 = x 1 or x 0 = x 2 Since g (x 2 ) = lim x x 2 g(x) g(x 2 ) x x 2 > 0, there is δ > 0 such that if 0 < x x 2 < δ, g(x) g(x 2) x x 0 > 0 Thus, if x (x 1, x 2 ) and x is close to x 2, then g(x) g(x 2 ) < 0, ie, g(x) < g(x 2 ) Thus, x 0 x 2 Since g (x 1 ) = lim x x 1 g(x) g(x 1 ) x x 1 > 0, there is δ > 0 such that if 0 < x x 1 < δ, g(x) g(x 2) x x 0 < 0 Thus, if x (x 1, x 2 ) and x is close to x 1, then g(x) g(x 2 ) < 0, ie, g(x) < g(x 2 ) Thus, x 0 x 2 So the proof is finished Remark The above theorem does not follow from the IVT for continuous functions since f may not be continuous (cf Exercise 284) From this theorem, we see that, if f does not satisfy the IVT, then it has no antiderivative 71