불연속인 도함수
모든 점에서 미분가능한데 도함수만 불연속인 함수입니다. 원점 근처를 확대해 도함수의 진동을 봅니다.
함수 자체는 원점에서 매끄럽게 가라앉습니다. 그런데 기울기는 진동이 멈추지 않습니다. 미분가능해도 도함수가 연속일 필요는 없습니다. 그래도 도함수는 중간값을 건너뛰지 않습니다.
예를 들면, ⚽ 공이 바닥에 닿을 때마다 점점 작게, 점점 빠르게 튀는 모습과 같아요. 공의 높이는 차분히 0으로 가라앉아요. 하지만 튀는 속도는 마지막까지 이리저리 흔들려요.
f ′ ( 0 ) = lim h → 0 h sin ( 1 / h ) = 0 f'(0) = \lim_{h \to 0} h\sin(1/h) = 0 f ′ ( 0 ) = h → 0 lim h sin ( 1/ h ) = 0 f ′ ( x ) = 2 x sin ( 1 / x ) − cos ( 1 / x ) ( x ≠ 0 ) , f ′ ( 1 2 π n ) = − 1 f'(x) = 2x\sin(1/x) - \cos(1/x) \quad (x \neq 0), \qquad f'\!\left(\tfrac{1}{2\pi n}\right) = -1 f ′ ( x ) = 2 x sin ( 1/ x ) − cos ( 1/ x ) ( x = 0 ) , f ′ ( 2 π n 1 ) = − 1 예제
f ( x ) = x 2 sin ( 1 / x ) f(x) = x^2\sin(1/x) f ( x ) = x 2 sin ( 1/ x ) x ≠ 0 x \ne 0 x = 0 f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0
f ( h ) − f ( 0 ) h = h sin 1 h , ∣ h sin 1 h ∣ ≤ ∣ h ∣ → 0 \frac{f(h) - f(0)}{h} = h\sin\frac{1}{h}, \qquad \left|h\sin\frac{1}{h}\right| \le |h| \to 0 h f ( h ) − f ( 0 ) = h sin h 1 , h sin h 1 ≤ ∣ h ∣ → 0 r = 0.1 r = 0.1 r = 0.1 h = 0.01 h = 0.01 h = 0.01 ≈ − 0.0051 \approx -0.0051 ≈ − 0.0051 증명 보기
도함수의 정의에 따라 f ′ ( 0 ) = lim h → 0 f ( h ) − f ( 0 ) h f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} f ′ ( 0 ) = lim h → 0 h f ( h ) − f ( 0 ) h ≠ 0 h \ne 0 h = 0 f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 h ≠ 0 h \ne 0 h = 0 f ( h ) = h 2 sin ( 1 / h ) f(h) = h^2\sin(1/h) f ( h ) = h 2 sin ( 1/ h ) h 2 sin ( 1 / h ) − 0 h = h sin ( 1 / h ) \frac{h^2\sin(1/h) - 0}{h} = h\sin(1/h) h h 2 s i n ( 1/ h ) − 0 = h sin ( 1/ h )
∣ sin ( 1 / h ) ∣ ≤ 1 |\sin(1/h)| \le 1 ∣ sin ( 1/ h ) ∣ ≤ 1 ∣ h sin ( 1 / h ) ∣ ≤ ∣ h ∣ |h\sin(1/h)| \le |h| ∣ h sin ( 1/ h ) ∣ ≤ ∣ h ∣
∣ h ∣ → 0 |h| \to 0 ∣ h ∣ → 0 h → 0 h \to 0 h → 0 h sin ( 1 / h ) → 0 h\sin(1/h) \to 0 h sin ( 1/ h ) → 0 f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0
진동하는 sin ( 1 / h ) \sin(1/h) sin ( 1/ h ) h h h
씬에서 h h h 0.0001 0.0001 0.0001 ∣ h sin ( 1 / h ) ∣ ≤ ∣ h ∣ |h\sin(1/h)| \le |h| ∣ h sin ( 1/ h ) ∣ ≤ ∣ h ∣
x ≠ 0 x \ne 0 x = 0 f ′ ( x ) = 2 x sin ( 1 / x ) − cos ( 1 / x ) f'(x) = 2x\sin(1/x) - \cos(1/x) f ′ ( x ) = 2 x sin ( 1/ x ) − cos ( 1/ x )
f ′ ( x ) = d d x [ x 2 sin 1 x ] = 2 x sin 1 x − cos 1 x ( x ≠ 0 ) f'(x) = \frac{d}{dx}\left[x^2\sin\frac{1}{x}\right] = 2x\sin\frac{1}{x} - \cos\frac{1}{x} \qquad (x \ne 0) f ′ ( x ) = d x d [ x 2 sin x 1 ] = 2 x sin x 1 − cos x 1 ( x = 0 ) r = 0.1 r = 0.1 r = 0.1 2 x sin ( 1 / x ) 2x\sin(1/x) 2 x sin ( 1/ x ) − cos ( 1 / x ) -\cos(1/x) − cos ( 1/ x ) 증명 보기
x ≠ 0 x \ne 0 x = 0 f ( x ) = x 2 ⋅ sin ( 1 / x ) f(x) = x^2 \cdot \sin(1/x) f ( x ) = x 2 ⋅ sin ( 1/ x ) f ′ ( x ) = 2 x ⋅ sin ( 1 / x ) + x 2 ⋅ d d x sin ( 1 / x ) f'(x) = 2x \cdot \sin(1/x) + x^2 \cdot \frac{d}{dx}\sin(1/x) f ′ ( x ) = 2 x ⋅ sin ( 1/ x ) + x 2 ⋅ d x d sin ( 1/ x )
연쇄 법칙으로 d d x sin ( 1 / x ) = cos ( 1 / x ) ⋅ d d x ( 1 / x ) = cos ( 1 / x ) ⋅ ( − 1 / x 2 ) \frac{d}{dx}\sin(1/x) = \cos(1/x) \cdot \frac{d}{dx}(1/x) = \cos(1/x) \cdot (-1/x^2) d x d sin ( 1/ x ) = cos ( 1/ x ) ⋅ d x d ( 1/ x ) = cos ( 1/ x ) ⋅ ( − 1/ x 2 ) x ≠ 0 x \ne 0 x = 0 1 / x 1/x 1/ x f ′ ( x ) = 2 x sin ( 1 / x ) + x 2 ⋅ cos ( 1 / x ) ⋅ ( − 1 / x 2 ) = 2 x sin ( 1 / x ) − cos ( 1 / x ) f'(x) = 2x\sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-1/x^2) = 2x\sin(1/x) - \cos(1/x) f ′ ( x ) = 2 x sin ( 1/ x ) + x 2 ⋅ cos ( 1/ x ) ⋅ ( − 1/ x 2 ) = 2 x sin ( 1/ x ) − cos ( 1/ x )
수치 확인: f ′ ( 0.5 ) = 2 ( 0.5 ) sin 2 − cos 2 ≈ 1.3254 f'(0.5) = 2(0.5)\sin 2 - \cos 2 \approx 1.3254 f ′ ( 0.5 ) = 2 ( 0.5 ) sin 2 − cos 2 ≈ 1.3254 x = 0 x = 0 x = 0 f ′ ( 0 ) f'(0) f ′ ( 0 )
f ′ f' f ′ x = 0 x = 0 x = 0 x n = 1 / ( 2 π n ) x_n = 1/(2\pi n) x n = 1/ ( 2 π n )
x n = 1 2 π n → 0 , f ′ ( x n ) = 2 x n sin ( 2 π n ) − cos ( 2 π n ) = − 1 ≠ 0 = f ′ ( 0 ) x_n = \frac{1}{2\pi n} \to 0, \qquad f'(x_n) = 2x_n\sin(2\pi n) - \cos(2\pi n) = -1 \ne 0 = f'(0) x n = 2 π n 1 → 0 , f ′ ( x n ) = 2 x n sin ( 2 π n ) − cos ( 2 π n ) = − 1 = 0 = f ′ ( 0 ) r = 0.01 r = 0.01 r = 0.01 h = 0.001 h = 0.001 h = 0.001 x n = 1 / ( 2 π n ) x_n = 1/(2\pi n) x n = 1/ ( 2 π n ) y = − 1 y = -1 y = − 1 증명 보기
x n = 1 / ( 2 π n ) x_n = 1/(2\pi n) x n = 1/ ( 2 π n ) x n ≠ 0 x_n \ne 0 x n = 0 x n → 0 x_n \to 0 x n → 0
문제 2의 공식에 대입하면 f ′ ( x n ) = 2 x n sin ( 1 / x n ) − cos ( 1 / x n ) = 2 x n sin ( 2 π n ) − cos ( 2 π n ) f'(x_n) = 2x_n\sin(1/x_n) - \cos(1/x_n) = 2x_n\sin(2\pi n) - \cos(2\pi n) f ′ ( x n ) = 2 x n sin ( 1/ x n ) − cos ( 1/ x n ) = 2 x n sin ( 2 π n ) − cos ( 2 π n )
sin ( 2 π n ) = 0 \sin(2\pi n) = 0 sin ( 2 π n ) = 0 cos ( 2 π n ) = 1 \cos(2\pi n) = 1 cos ( 2 π n ) = 1 f ′ ( x n ) = 0 − 1 = − 1 f'(x_n) = 0 - 1 = -1 f ′ ( x n ) = 0 − 1 = − 1 n n n
따라서 f ′ ( x n ) → − 1 f'(x_n) \to -1 f ′ ( x n ) → − 1 f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 x n → 0 x_n \to 0 x n → 0 f ′ ( x n ) ↛ f ′ ( 0 ) f'(x_n) \not\to f'(0) f ′ ( x n ) → f ′ ( 0 ) f ′ f' f ′ 0 0 0 lim x → 0 f ′ ( x ) \lim_{x \to 0} f'(x) lim x → 0 f ′ ( x )
덧붙임: 도함수가 불연속이어도 중간값 성질은 잃지 않습니다 — 다르부 정리(Darboux's theorem)에 의해 f ′ f' f ′
10. 미분