앞 강의에서 다항함수는 항별로 미분했습니다. 그러나 함수가 곱이나 몫으로 얽히거나, 함수 안에 함수가 들어 있으면 항별 미분만으로는 되지 않습니다. 이 강의는 복잡한 함수를 구조별로 쪼개어 미분하는 세 법칙, 곱의 법칙·몫의 법칙·연쇄법칙을 유도하고 손에 익힙니다. 미분에서 실수의 대부분은 "구조를 잘못 보거나 안쪽 미분을 빠뜨리는 것"입니다.
곱의 법칙으로 두 함수의 곱을 미분할 수 있습니다.
몫의 법칙으로 분수 꼴 함수를 미분할 수 있습니다.
연쇄법칙으로 합성함수를 미분할 수 있습니다.
함수를 곱·몫·합성 중 어느 구조로 볼지 먼저 판단할 수 있습니다.
세 법칙을 겹쳐 쓰는 복합 함수를 미분할 수 있습니다.
기호
읽는 법
뜻
( f g ) ′ ( f g ) ′ 에프지 프라임
두 함수 곱의 도함수
( f g ) ′ ( g f ) ′ 에프 오버 지 프라임
두 함수 몫의 도함수
( f ∘ g ) ′ ( f ∘ g ) ′ 에프 서클 지 프라임
합성함수 f ( g ( x ) ) f ( g ( x ) )
u = g ( x ) u = g ( x ) 유는 지엑스
연쇄법칙에서 안쪽 함수(중간 변수)
\dfrac{dy}{du}\dfrac{du}
디와이 디유 곱하기 디유 디엑스
연쇄법칙의 라이프니츠 표기
두 함수의 곱을 미분할 때, 각각을 따로 미분해 곱하면 틀립니다. 올바른 규칙은 이렇습니다.
곱의 법칙. d d x [ f ( x ) g ( x ) ] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) d x d [ f ( x ) g ( x ) ] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x )
유도는 정의에서 나옵니다. 분자 f ( x + h ) g ( x + h ) − f ( x ) g ( x ) f ( x + h ) g ( x + h ) − f ( x ) g ( x ) − f ( x ) g ( x + h ) + f ( x ) g ( x + h ) − f ( x ) g ( x + h ) + f ( x ) g ( x + h )
[ f ( x + h ) − f ( x ) ] g ( x + h ) + f ( x ) [ g ( x + h ) − g ( x ) ] h h [ f ( x + h ) − f ( x ) ] g ( x + h ) + f ( x ) [ g ( x + h ) − g ( x ) ]
로 갈라지고, h → 0 h → 0 f ′ ( x ) g ( x ) f ′ ( x ) g ( x ) f ( x ) g ′ ( x ) f ( x ) g ′ ( x )
예를 들어 y = ( x 2 + 1 ) ( x 3 − x ) y = ( x 2 + 1 ) ( x 3 − x )
y ′ = ( x 2 + 1 ) ′ ( x 3 − x ) + ( x 2 + 1 ) ( x 3 − x ) ′ = 2 x ( x 3 − x ) + ( x 2 + 1 ) ( 3 x 2 − 1 ) y ′ = ( x 2 + 1 ) ′ ( x 3 − x ) + ( x 2 + 1 ) ( x 3 − x ) ′ = 2 x ( x 3 − x ) + ( x 2 + 1 ) ( 3 x 2 − 1 )
전개하면 2 x 4 − 2 x 2 + 3 x 4 − x 2 + 3 x 2 − 1 = 5 x 4 + 0 x 2 − 1 = 5 x 4 − 1 2 x 4 − 2 x 2 + 3 x 4 − x 2 + 3 x 2 − 1 = 5 x 4 + 0 x 2 − 1 = 5 x 4 − 1 y = x 5 − x 3 + x 3 − x = x 5 − x y = x 5 − x 3 + x 3 − x = x 5 − x 5 x 4 − 1 5 x 4 − 1
문제 1. (기초) y = x 2 ( x + 1 ) y = x 2 ( x + 1 ) 답. 3 x 2 + 2 x 3 x 2 + 2 x 풀이. y ′ = ( x 2 ) ′ ( x + 1 ) + x 2 ( x + 1 ) ′ = 2 x ( x + 1 ) + x 2 ⋅ 1 = 2 x 2 + 2 x + x 2 = 3 x 2 + 2 x y ′ = ( x 2 ) ′ ( x + 1 ) + x 2 ( x + 1 ) ′ = 2 x ( x + 1 ) + x 2 ⋅ 1 = 2 x 2 + 2 x + x 2 = 3 x 2 + 2 x y = x 3 + x 2 → 3 x 2 + 2 x y = x 3 + x 2 → 3 x 2 + 2 x
문제 2. (표준) y = ( 2 x − 1 ) ( x 2 + 3 ) y = ( 2 x − 1 ) ( x 2 + 3 ) 답. 6 x 2 − 2 x + 6 6 x 2 − 2 x + 6 풀이. y ′ = ( 2 x − 1 ) ′ ( x 2 + 3 ) + ( 2 x − 1 ) ( x 2 + 3 ) ′ = 2 ( x 2 + 3 ) + ( 2 x − 1 ) ( 2 x ) = 2 x 2 + 6 + 4 x 2 − 2 x = 6 x 2 − 2 x + 6 y ′ = ( 2 x − 1 ) ′ ( x 2 + 3 ) + ( 2 x − 1 ) ( x 2 + 3 ) ′ = 2 ( x 2 + 3 ) + ( 2 x − 1 ) ( 2 x ) = 2 x 2 + 6 + 4 x 2 − 2 x = 6 x 2 − 2 x + 6
문제 3. (표준) y = x x ( x + 1 ) y = x x ( x + 1 ) x > 0 x > 0 답. 5 2 x 3 / 2 + 3 2 x 1 / 2 2 5 x 3 / 2 + 2 3 x 1 / 2 풀이. x x = x 3 / 2 x x = x 3 / 2 y = x 3 / 2 ( x + 1 ) y = x 3 / 2 ( x + 1 ) y ′ = 3 2 x 1 / 2 ( x + 1 ) + x 3 / 2 ⋅ 1 = 3 2 x 3 / 2 + 3 2 x 1 / 2 + x 3 / 2 = 5 2 x 3 / 2 + 3 2 x 1 / 2 y ′ = 2 3 x 1 / 2 ( x + 1 ) + x 3 / 2 ⋅ 1 = 2 3 x 3 / 2 + 2 3 x 1 / 2 + x 3 / 2 = 2 5 x 3 / 2 + 2 3 x 1 / 2
문제 4. (심화) y = ( x 2 + 1 ) ( x 2 − 1 ) ( x ) y = ( x 2 + 1 ) ( x 2 − 1 ) ( x ) x = 1 x = 1 답. 4 4 풀이. 세 인수의 곱은 전개가 편합니다. ( x 2 + 1 ) ( x 2 − 1 ) = x 4 − 1 ( x 2 + 1 ) ( x 2 − 1 ) = x 4 − 1 y = x ( x 4 − 1 ) = x 5 − x y = x ( x 4 − 1 ) = x 5 − x y ′ = 5 x 4 − 1 y ′ = 5 x 4 − 1 x = 1 x = 1 5 ⋅ 1 − 1 = 4 5 ⋅ 1 − 1 = 4
분수 꼴 함수는 다음 규칙으로 미분합니다. 부호와 순서가 핵심이라 외울 때 "아래 제곱분의, 위미분 곱하기 아래 빼기 위 곱하기 아래미분"으로 기억합니다.
몫의 법칙. \dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)\,g(x)-f(x)\,g'(x)}
분자에서 순서를 바꾸면 부호가 뒤집혀 틀리므로 주의합니다. 예를 들어 y = x x 2 + 1 y = x 2 + 1 x
y ′ = ( x ) ′ ( x 2 + 1 ) − x ( x 2 + 1 ) ′ ( x 2 + 1 ) 2 = 1 ⋅ ( x 2 + 1 ) − x ⋅ 2 x ( x 2 + 1 ) 2 = x 2 + 1 − 2 x 2 ( x 2 + 1 ) 2 = 1 − x 2 ( x 2 + 1 ) 2 y ′ = ( x 2 + 1 ) 2 ( x ) ′ ( x 2 + 1 ) − x ( x 2 + 1 ) ′ = ( x 2 + 1 ) 2 1 ⋅ ( x 2 + 1 ) − x ⋅ 2 x = ( x 2 + 1 ) 2 x 2 + 1 − 2 x 2 = ( x 2 + 1 ) 2 1 − x 2
입니다.
문제 1. (기초) y = 1 x + 1 y = x + 1 1 답. − 1 ( x + 1 ) 2 − ( x + 1 ) 2 1 풀이. y ′ = 0 ⋅ ( x + 1 ) − 1 ⋅ 1 ( x + 1 ) 2 = − 1 ( x + 1 ) 2 y ′ = ( x + 1 ) 2 0 ⋅ ( x + 1 ) − 1 ⋅ 1 = ( x + 1 ) 2 − 1
문제 2. (표준) y = x − 1 x + 1 y = x + 1 x − 1 답. 2 ( x + 1 ) 2 ( x + 1 ) 2 2 풀이. y ′ = ( 1 ) ( x + 1 ) − ( x − 1 ) ( 1 ) ( x + 1 ) 2 = ( x + 1 ) − ( x − 1 ) ( x + 1 ) 2 = 2 ( x + 1 ) 2 y ′ = ( x + 1 ) 2 ( 1 ) ( x + 1 ) − ( x − 1 ) ( 1 ) = ( x + 1 ) 2 ( x + 1 ) − ( x − 1 ) = ( x + 1 ) 2 2
문제 3. (표준) y = x 2 x − 1 y = x − 1 x 2 답. x 2 − 2 x ( x − 1 ) 2 ( x − 1 ) 2 x 2 − 2 x 풀이. y ′ = ( 2 x ) ( x − 1 ) − x 2 ( 1 ) ( x − 1 ) 2 = 2 x 2 − 2 x − x 2 ( x − 1 ) 2 = x 2 − 2 x ( x − 1 ) 2 y ′ = ( x − 1 ) 2 ( 2 x ) ( x − 1 ) − x 2 ( 1 ) = ( x − 1 ) 2 2 x 2 − 2 x − x 2 = ( x − 1 ) 2 x 2 − 2 x
문제 4. (심화) y = x 2 + 1 x 2 − 1 y = x 2 − 1 x 2 + 1 x = 2 x = 2 답. − 8 9 − 9 8 풀이. y ′ = ( 2 x ) ( x 2 − 1 ) − ( x 2 + 1 ) ( 2 x ) ( x 2 − 1 ) 2 = 2 x [ ( x 2 − 1 ) − ( x 2 + 1 ) ] ( x 2 − 1 ) 2 = 2 x ⋅ ( − 2 ) ( x 2 − 1 ) 2 = − 4 x ( x 2 − 1 ) 2 y ′ = ( x 2 − 1 ) 2 ( 2 x ) ( x 2 − 1 ) − ( x 2 + 1 ) ( 2 x ) = ( x 2 − 1 ) 2 2 x [ ( x 2 − 1 ) − ( x 2 + 1 ) ] = ( x 2 − 1 ) 2 2 x ⋅ ( − 2 ) = ( x 2 − 1 ) 2 − 4 x x = 2 x = 2 ( 4 − 1 ) 2 = 9 ( 4 − 1 ) 2 = 9 − 4 ⋅ 2 9 = − 8 9 9 − 4 ⋅ 2 = − 9 8
함수 안에 함수가 들어 있는 합성함수는 연쇄법칙으로 미분합니다. 이것이 미분에서 가장 자주 쓰이면서 가장 자주 틀리는 규칙입니다.
연쇄법칙. d d x f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x ) d x d f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x )
라이프니츠 표기로는 u = g ( x ) u = g ( x ) d y d x = d y d u ⋅ d u d x d x d y = d u d y ⋅ d x d u
직관은 "바깥을 먼저 미분하되 안은 그대로 두고, 마지막에 안쪽의 미분을 곱한다"입니다. 안쪽 미분 g ′ ( x ) g ′ ( x )
예를 들어 y = ( x 2 + 1 ) 3 y = ( x 2 + 1 ) 3 u = x 2 + 1 u = x 2 + 1
y ′ = 3 ( x 2 + 1 ) 2 ⋅ ( x 2 + 1 ) ′ = 3 ( x 2 + 1 ) 2 ⋅ 2 x = 6 x ( x 2 + 1 ) 2 y ′ = 3 ( x 2 + 1 ) 2 ⋅ ( x 2 + 1 ) ′ = 3 ( x 2 + 1 ) 2 ⋅ 2 x = 6 x ( x 2 + 1 ) 2
또 y = 3 x + 1 = ( 3 x + 1 ) 1 / 2 y = 3 x + 1 = ( 3 x + 1 ) 1 / 2
y ′ = 1 2 ( 3 x + 1 ) − 1 / 2 ⋅ 3 = 3 2 3 x + 1 y ′ = 2 1 ( 3 x + 1 ) − 1 / 2 ⋅ 3 = 2 3 x + 1 3
입니다. 안쪽 3 x + 1 3 x + 1 3 3
문제 1. (기초) y = ( 2 x + 1 ) 4 y = ( 2 x + 1 ) 4 답. 8 ( 2 x + 1 ) 3 8 ( 2 x + 1 ) 3 풀이. 바깥은 네제곱, 안쪽 미분은 2 2 y ′ = 4 ( 2 x + 1 ) 3 ⋅ 2 = 8 ( 2 x + 1 ) 3 y ′ = 4 ( 2 x + 1 ) 3 ⋅ 2 = 8 ( 2 x + 1 ) 3
문제 2. (기초) y = ( x 3 − x ) 5 y = ( x 3 − x ) 5 답. 5 ( x 3 − x ) 4 ( 3 x 2 − 1 ) 5 ( x 3 − x ) 4 ( 3 x 2 − 1 ) 풀이. y ′ = 5 ( x 3 − x ) 4 ⋅ ( x 3 − x ) ′ = 5 ( x 3 − x ) 4 ( 3 x 2 − 1 ) y ′ = 5 ( x 3 − x ) 4 ⋅ ( x 3 − x ) ′ = 5 ( x 3 − x ) 4 ( 3 x 2 − 1 )
문제 3. (표준) y = x 2 + 4 y = x 2 + 4 답. x x 2 + 4 x 2 + 4 x 풀이. y = ( x 2 + 4 ) 1 / 2 y = ( x 2 + 4 ) 1 / 2 y ′ = 1 2 ( x 2 + 4 ) − 1 / 2 ⋅ 2 x = x x 2 + 4 y ′ = 2 1 ( x 2 + 4 ) − 1 / 2 ⋅ 2 x = x 2 + 4 x
문제 4. (표준) y = 1 ( x 2 + 1 ) 2 y = ( x 2 + 1 ) 2 1 답. − 4 x ( x 2 + 1 ) 3 − ( x 2 + 1 ) 3 4 x 풀이. y = ( x 2 + 1 ) − 2 y = ( x 2 + 1 ) − 2 y ′ = − 2 ( x 2 + 1 ) − 3 ⋅ 2 x = − 4 x ( x 2 + 1 ) 3 y ′ = − 2 ( x 2 + 1 ) − 3 ⋅ 2 x = − ( x 2 + 1 ) 3 4 x
문제 5. (심화) y = ( ( x + 1 ) 2 + 1 ) 3 y = ( ( x + 1 ) 2 + 1 ) 3 답. 6 ( ( x + 1 ) 2 + 1 ) 2 ( x + 1 ) 6 ( ( x + 1 ) 2 + 1 ) 2 ( x + 1 ) 풀이. 연쇄를 두 번 씁니다. 바깥 세제곱을 미분하면 3 ( ( x + 1 ) 2 + 1 ) 2 3 ( ( x + 1 ) 2 + 1 ) 2 ( x + 1 ) 2 + 1 ( x + 1 ) 2 + 1 2 ( x + 1 ) ⋅ 1 = 2 ( x + 1 ) 2 ( x + 1 ) ⋅ 1 = 2 ( x + 1 ) 3 ( ( x + 1 ) 2 + 1 ) 2 ⋅ 2 ( x + 1 ) = 6 ( ( x + 1 ) 2 + 1 ) 2 ( x + 1 ) 3 ( ( x + 1 ) 2 + 1 ) 2 ⋅ 2 ( x + 1 ) = 6 ( ( x + 1 ) 2 + 1 ) 2 ( x + 1 )
실제 함수는 곱·몫·합성이 뒤섞여 있습니다. 먼저 전체가 어떤 구조인지(가장 바깥이 곱인가, 몫인가, 합성인가) 판단하고, 필요한 곳에서 연쇄법칙을 부품처럼 끼워 넣습니다.
예를 들어 y = x 2 x + 1 y = x 2 x + 1 x + 1 x + 1
y ′ = ( x 2 ) ′ x + 1 + x 2 ⋅ ( x + 1 ) ′ = 2 x x + 1 + x 2 ⋅ 1 2 x + 1 y ′ = ( x 2 ) ′ x + 1 + x 2 ⋅ ( x + 1 ) ′ = 2 x x + 1 + x 2 ⋅ 2 x + 1 1
통분하면 4 x ( x + 1 ) + x 2 2 x + 1 = 5 x 2 + 4 x 2 x + 1 2 x + 1 4 x ( x + 1 ) + x 2 = 2 x + 1 5 x 2 + 4 x
또 y = ( 2 x + 1 ) 3 x y = x ( 2 x + 1 ) 3
y ′ = [ 3 ( 2 x + 1 ) 2 ⋅ 2 ] x − ( 2 x + 1 ) 3 ⋅ 1 x 2 = 6 x ( 2 x + 1 ) 2 − ( 2 x + 1 ) 3 x 2 y ′ = x 2 [ 3 ( 2 x + 1 ) 2 ⋅ 2 ] x − ( 2 x + 1 ) 3 ⋅ 1 = x 2 6 x ( 2 x + 1 ) 2 − ( 2 x + 1 ) 3
공통인수 ( 2 x + 1 ) 2 ( 2 x + 1 ) 2 ( 2 x + 1 ) 2 [ 6 x − ( 2 x + 1 ) ] x 2 = ( 2 x + 1 ) 2 ( 4 x − 1 ) x 2 x 2 ( 2 x + 1 ) 2 [ 6 x − ( 2 x + 1 ) ] = x 2 ( 2 x + 1 ) 2 ( 4 x − 1 )
문제 1. (표준) y = x ( x + 2 ) 3 y = x ( x + 2 ) 3 답. ( x + 2 ) 2 ( 4 x + 2 ) ( x + 2 ) 2 ( 4 x + 2 ) 풀이. 곱의 법칙과 연쇄를 함께 씁니다. y ′ = 1 ⋅ ( x + 2 ) 3 + x ⋅ 3 ( x + 2 ) 2 = ( x + 2 ) 3 + 3 x ( x + 2 ) 2 y ′ = 1 ⋅ ( x + 2 ) 3 + x ⋅ 3 ( x + 2 ) 2 = ( x + 2 ) 3 + 3 x ( x + 2 ) 2 ( x + 2 ) 2 ( x + 2 ) 2 ( x + 2 ) 2 [ ( x + 2 ) + 3 x ] = ( x + 2 ) 2 ( 4 x + 2 ) ( x + 2 ) 2 [ ( x + 2 ) + 3 x ] = ( x + 2 ) 2 ( 4 x + 2 )
문제 2. (표준) y = x x + 1 y = x + 1 x x > 0 x > 0 답. 1 − x 2 x ( x + 1 ) 2 2 x ( x + 1 ) 2 1 − x 풀이. 몫의 법칙을 씁니다. y ′ = 1 2 x ( x + 1 ) − x ⋅ 1 ( x + 1 ) 2 y ′ = ( x + 1 ) 2 2 x 1 ( x + 1 ) − x ⋅ 1 2 x 2 x ( x + 1 ) − 2 x 2 x = 1 − x 2 x 2 x ( x + 1 ) − 2 x = 2 x 1 − x y ′ = 1 − x 2 x ( x + 1 ) 2 y ′ = 2 x ( x + 1 ) 2 1 − x
문제 3. (심화) y = x − 1 x + 1 y = x + 1 x − 1 답. 1 ( x + 1 ) 2 x + 1 x − 1 ( x + 1 ) 2 1 x − 1 x + 1 풀이. u = x − 1 x + 1 u = x + 1 x − 1 y = u 1 / 2 y = u 1 / 2 y ′ = 1 2 u − 1 / 2 ⋅ u ′ y ′ = 2 1 u − 1 / 2 ⋅ u ′ u ′ = ( 1 ) ( x + 1 ) − ( x − 1 ) ( 1 ) ( x + 1 ) 2 = 2 ( x + 1 ) 2 u ′ = ( x + 1 ) 2 ( 1 ) ( x + 1 ) − ( x − 1 ) ( 1 ) = ( x + 1 ) 2 2 y ′ = 1 2 ( x − 1 x + 1 ) − 1 / 2 ⋅ 2 ( x + 1 ) 2 = 1 ( x + 1 ) 2 x + 1 x − 1 y ′ = 2 1 ( x + 1 x − 1 ) − 1 / 2 ⋅ ( x + 1 ) 2 2 = ( x + 1 ) 2 1 x − 1 x + 1
곱: ( f g ) ′ = f ′ g + f g ′ ( f g ) ′ = f ′ g + f g ′
몫: ( f g ) ′ = f ′ g − f g ′ g 2 ( g f ) ′ = g 2 f ′ g − f g ′
연쇄: d d x f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x ) d x d f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x )
거듭제곱+연쇄: d d x [ g ( x ) ] n = n [ g ( x ) ] n − 1 g ′ ( x ) d x d [ g ( x ) ] n = n [ g ( x ) ] n − 1 g ′ ( x )
근호: d d x g ( x ) = g ′ ( x ) 2 g ( x ) d x d g ( x ) = 2 g ( x ) g ′ ( x )
복합식은 가장 바깥 구조부터 판단하고, 안쪽에 연쇄를 부품으로 끼운다.
곱을 전개할 수 있으면 전개해 검산하면 실수를 잡는다.
문제 1. (기초) y = ( 3 x + 2 ) 5 y = ( 3 x + 2 ) 5 답. 15 ( 3 x + 2 ) 4 1 5 ( 3 x + 2 ) 4 풀이. y ′ = 5 ( 3 x + 2 ) 4 ⋅ 3 = 15 ( 3 x + 2 ) 4 y ′ = 5 ( 3 x + 2 ) 4 ⋅ 3 = 1 5 ( 3 x + 2 ) 4
문제 2. (기초) y = ( x 2 − 2 x ) ( x + 1 ) y = ( x 2 − 2 x ) ( x + 1 ) 답. 3 x 2 − 2 x − 2 3 x 2 − 2 x − 2 풀이. y ′ = ( 2 x − 2 ) ( x + 1 ) + ( x 2 − 2 x ) ( 1 ) = 2 x 2 − 2 + 2 x − 2 x + x 2 − 2 x = 3 x 2 − 2 x − 2 y ′ = ( 2 x − 2 ) ( x + 1 ) + ( x 2 − 2 x ) ( 1 ) = 2 x 2 − 2 + 2 x − 2 x + x 2 − 2 x = 3 x 2 − 2 x − 2 y = x 3 − x 2 − 2 x → 3 x 2 − 2 x − 2 y = x 3 − x 2 − 2 x → 3 x 2 − 2 x − 2
문제 3. (표준) y = 2 x + 1 x − 3 y = x − 3 2 x + 1 답. − 7 ( x − 3 ) 2 − ( x − 3 ) 2 7 풀이. y ′ = 2 ( x − 3 ) − ( 2 x + 1 ) ( 1 ) ( x − 3 ) 2 = 2 x − 6 − 2 x − 1 ( x − 3 ) 2 = − 7 ( x − 3 ) 2 y ′ = ( x − 3 ) 2 2 ( x − 3 ) − ( 2 x + 1 ) ( 1 ) = ( x − 3 ) 2 2 x − 6 − 2 x − 1 = ( x − 3 ) 2 − 7
문제 4. (표준) y = x 2 − 2 x + 2 y = x 2 − 2 x + 2 답. x − 1 x 2 − 2 x + 2 x 2 − 2 x + 2 x − 1 풀이. y ′ = ( x 2 − 2 x + 2 ) ′ 2 x 2 − 2 x + 2 = 2 x − 2 2 x 2 − 2 x + 2 = x − 1 x 2 − 2 x + 2 y ′ = 2 x 2 − 2 x + 2 ( x 2 − 2 x + 2 ) ′ = 2 x 2 − 2 x + 2 2 x − 2 = x 2 − 2 x + 2 x − 1
문제 5. (표준) y = x 2 ( 2 x − 1 ) 3 y = x 2 ( 2 x − 1 ) 3 답. 2 x ( 2 x − 1 ) 2 ( 5 x − 1 ) 2 x ( 2 x − 1 ) 2 ( 5 x − 1 ) 풀이. y ′ = 2 x ( 2 x − 1 ) 3 + x 2 ⋅ 3 ( 2 x − 1 ) 2 ⋅ 2 = 2 x ( 2 x − 1 ) 3 + 6 x 2 ( 2 x − 1 ) 2 y ′ = 2 x ( 2 x − 1 ) 3 + x 2 ⋅ 3 ( 2 x − 1 ) 2 ⋅ 2 = 2 x ( 2 x − 1 ) 3 + 6 x 2 ( 2 x − 1 ) 2 2 x ( 2 x − 1 ) 2 2 x ( 2 x − 1 ) 2 2 x ( 2 x − 1 ) 2 [ ( 2 x − 1 ) + 3 x ] = 2 x ( 2 x − 1 ) 2 ( 5 x − 1 ) 2 x ( 2 x − 1 ) 2 [ ( 2 x − 1 ) + 3 x ] = 2 x ( 2 x − 1 ) 2 ( 5 x − 1 )
문제 6. (표준) y = x x 2 + 1 y = x 2 + 1 x 답. 1 ( x 2 + 1 ) 3 / 2 ( x 2 + 1 ) 3 / 2 1 풀이. 몫의 법칙을 씁니다. 분모 g = x 2 + 1 g = x 2 + 1 x x 2 + 1 x 2 + 1 x y ′ = 1 ⋅ x 2 + 1 − x ⋅ x x 2 + 1 x 2 + 1 y ′ = x 2 + 1 1 ⋅ x 2 + 1 − x ⋅ x 2 + 1 x x 2 + 1 x 2 + 1 ( x 2 + 1 ) − x 2 x 2 + 1 = 1 x 2 + 1 x 2 + 1 ( x 2 + 1 ) − x 2 = x 2 + 1 1 y ′ = 1 ( x 2 + 1 ) 3 / 2 y ′ = ( x 2 + 1 ) 3 / 2 1
문제 7. (표준) y = ( x 2 + 1 ) 3 ( x − 1 ) 2 y = ( x 2 + 1 ) 3 ( x − 1 ) 2 x = 1 x = 1 답. 0 0 풀이. y ′ = 3 ( x 2 + 1 ) 2 ⋅ 2 x ⋅ ( x − 1 ) 2 + ( x 2 + 1 ) 3 ⋅ 2 ( x − 1 ) y ′ = 3 ( x 2 + 1 ) 2 ⋅ 2 x ⋅ ( x − 1 ) 2 + ( x 2 + 1 ) 3 ⋅ 2 ( x − 1 ) x = 1 x = 1 ( x − 1 ) ( x − 1 ) y ′ ( 1 ) = 0 y ′ ( 1 ) = 0
문제 8. (심화) y = ( x − 1 x + 1 ) 3 y = ( x + 1 x − 1 ) 3 답. 6 ( x − 1 ) 2 ( x + 1 ) 4 ( x + 1 ) 4 6 ( x − 1 ) 2 풀이. u = x − 1 x + 1 u = x + 1 x − 1 y = u 3 y = u 3 y ′ = 3 u 2 u ′ y ′ = 3 u 2 u ′ u ′ = ( x + 1 ) − ( x − 1 ) ( x + 1 ) 2 = 2 ( x + 1 ) 2 u ′ = ( x + 1 ) 2 ( x + 1 ) − ( x − 1 ) = ( x + 1 ) 2 2 y ′ = 3 ( x − 1 x + 1 ) 2 ⋅ 2 ( x + 1 ) 2 = 6 ( x − 1 ) 2 ( x + 1 ) 4 y ′ = 3 ( x + 1 x − 1 ) 2 ⋅ ( x + 1 ) 2 2 = ( x + 1 ) 4 6 ( x − 1 ) 2
문제 9. (심화) y = x + x y = x + x x > 0 x > 0 답. 1 + 1 2 x 2 x + x 2 x + x 1 + 2 x 1 풀이. 바깥 근호를 미분하면 ( x + x ) ′ 2 x + x 2 x + x ( x + x ) ′ x + x x + x 1 + 1 2 x 1 + 2 x 1 y ′ = 1 + 1 2 x 2 x + x y ′ = 2 x + x 1 + 2 x 1
문제 10. (심화) y = ( x + 1 ) 2 ( x − 1 ) 3 y = ( x − 1 ) 3 ( x + 1 ) 2 답. − ( x + 1 ) ( x + 5 ) ( x − 1 ) 4 − ( x − 1 ) 4 ( x + 1 ) ( x + 5 ) 풀이. 몫의 법칙을 씁니다. 분자 f = ( x + 1 ) 2 f = ( x + 1 ) 2 f ′ = 2 ( x + 1 ) f ′ = 2 ( x + 1 ) g = ( x − 1 ) 3 g = ( x − 1 ) 3 g ′ = 3 ( x − 1 ) 2 g ′ = 3 ( x − 1 ) 2 y ′ = 2 ( x + 1 ) ( x − 1 ) 3 − ( x + 1 ) 2 ⋅ 3 ( x − 1 ) 2 ( x − 1 ) 6 y ′ = ( x − 1 ) 6 2 ( x + 1 ) ( x − 1 ) 3 − ( x + 1 ) 2 ⋅ 3 ( x − 1 ) 2 ( x + 1 ) ( x − 1 ) 2 ( x + 1 ) ( x − 1 ) 2 ( x + 1 ) ( x − 1 ) 2 [ 2 ( x − 1 ) − 3 ( x + 1 ) ] = ( x + 1 ) ( x − 1 ) 2 ( − x − 5 ) ( x + 1 ) ( x − 1 ) 2 [ 2 ( x − 1 ) − 3 ( x + 1 ) ] = ( x + 1 ) ( x − 1 ) 2 ( − x − 5 ) ( x − 1 ) 6 ( x − 1 ) 6 y ′ = − ( x + 1 ) ( x + 5 ) ( x − 1 ) 4 y ′ = ( x − 1 ) 4 − ( x + 1 ) ( x + 5 )
문제 11. (심화) y = x 1 − x 2 y = x 1 − x 2 x = 1 2 x = 2 1 답. 1 3 3 1 풀이. y ′ = 1 − x 2 + x ⋅ − 2 x 2 1 − x 2 = 1 − x 2 − x 2 1 − x 2 y ′ = 1 − x 2 + x ⋅ 2 1 − x 2 − 2 x = 1 − x 2 − 1 − x 2 x 2 ( 1 − x 2 ) − x 2 1 − x 2 = 1 − 2 x 2 1 − x 2 1 − x 2 ( 1 − x 2 ) − x 2 = 1 − x 2 1 − 2 x 2 x = 1 2 x = 2 1 1 − 1 2 1 − 1 4 = 1 / 2 3 / 2 = 1 3 1 − 4 1 1 − 2 1 = 3 / 2 1 / 2 = 3 1
문제 12. (심화) y = ( 2 x − 1 ) 3 ( 3 x + 1 ) 2 y = ( 2 x − 1 ) 3 ( 3 x + 1 ) 2 답. 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) ( 5 x ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) ( 5 x ) 30 x ( 2 x − 1 ) 2 ( 3 x + 1 ) 3 0 x ( 2 x − 1 ) 2 ( 3 x + 1 ) 풀이. y ′ = 3 ( 2 x − 1 ) 2 ⋅ 2 ⋅ ( 3 x + 1 ) 2 + ( 2 x − 1 ) 3 ⋅ 2 ( 3 x + 1 ) ⋅ 3 y ′ = 3 ( 2 x − 1 ) 2 ⋅ 2 ⋅ ( 3 x + 1 ) 2 + ( 2 x − 1 ) 3 ⋅ 2 ( 3 x + 1 ) ⋅ 3 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) 2 + 6 ( 2 x − 1 ) 3 ( 3 x + 1 ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) 2 + 6 ( 2 x − 1 ) 3 ( 3 x + 1 ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) [ ( 3 x + 1 ) + ( 2 x − 1 ) ] = 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) ( 5 x ) = 30 x ( 2 x − 1 ) 2 ( 3 x + 1 ) 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) [ ( 3 x + 1 ) + ( 2 x − 1 ) ] = 6 ( 2 x − 1 ) 2 ( 3 x + 1 ) ( 5 x ) = 3 0 x ( 2 x − 1 ) 2 ( 3 x + 1 )
곱의 법칙 ( f g ) ′ = f ′ g + f g ′ ( f g ) ′ = f ′ g + f g ′
y = x x 2 + 1 y = x 2 + 1 x 1 − x 2 ( x 2 + 1 ) 2 ( x 2 + 1 ) 2 1 − x 2 연쇄법칙에서 안쪽 함수의 미분을 왜 곱해야 하는지 설명할 수 있는가?
y = ( x 2 + 3 ) 4 y = ( x 2 + 3 ) 4 8 x ( x 2 + 3 ) 3 8 x ( x 2 + 3 ) 3 y = 2 x + 3 y = 2 x + 3 1 2 x + 3 2 x + 3 1 곱·몫·합성이 섞인 식에서 가장 바깥 구조를 먼저 판단할 수 있는가?
y = x 2 ( x − 1 ) 3 y = x 2 ( x − 1 ) 3 x ( x − 1 ) 2 ( 5 x − 2 ) x ( x − 1 ) 2 ( 5 x − 2 ) 곱을 전개해 검산하는 습관을 갖고 있는가?