지금까지는 y = f ( x ) y = f ( x ) y y x x x 2 + y 2 = 1 x 2 + y 2 = 1 y y y y x x
음함수 관계식을 양변 미분해 d y d x d x d y
y y d y d x d x d y 음함수로 주어진 곡선의 접선을 구할 수 있습니다.
역함수의 도함수 공식을 유도하고 적용할 수 있습니다.
로그미분으로 복잡한 곱·몫·거듭제곱을 미분할 수 있습니다.
기호
읽는 법
뜻
\dfrac{dy}
디와이 디엑스
x x y y
d d x ( ⋅ ) d x d ( ⋅ ) 디 디엑스
양변을 x x
f^
에프 인버스
f f
( f − 1 ) ′ ( b ) ( f − 1 ) ′ ( b ) 에프 인버스 프라임 비
역함수의 미분계수
y ′ y ′ 와이 프라임
d y d x d x d y
x x y y F ( x , y ) = 0 F ( x , y ) = 0 y y x x
y y x x x x y y d y d x d x d y
예를 들어 y 2 y 2 x x y y 2 y ⋅ d y d x 2 y ⋅ d x d y y y d y d x d x d y y 3 y 3 3 y 2 d y d x 3 y 2 d x d y d y d x d x d y
원 x 2 + y 2 = 1 x 2 + y 2 = 1 x x
2 x + 2 y d y d x = 0 2 x + 2 y d x d y = 0
이고, d y d x d x d y
d y d x = − x y d x d y = − y x
입니다. y y ± 1 − x 2 ± 1 − x 2
문제 1. (기초) x 2 + y 2 = 25 x 2 + y 2 = 2 5 d y d x d x d y 답. − x y − y x 풀이. 양변을 미분하면 2 x + 2 y d y d x = 0 2 x + 2 y d x d y = 0 d y d x = − x y d x d y = − y x
문제 2. (기초) x y = 4 x y = 4 d y d x d x d y 답. − y x − x y 풀이. 좌변에 곱의 법칙을 씁니다. d d x ( x y ) = 1 ⋅ y + x d y d x d x d ( x y ) = 1 ⋅ y + x d x d y y + x d y d x = 0 y + x d x d y = 0 d y d x = − y x d x d y = − x y
문제 3. (표준) x 2 + x y + y 2 = 7 x 2 + x y + y 2 = 7 d y d x d x d y 답. − 2 x + y x + 2 y − x + 2 y 2 x + y 풀이. 양변을 미분하면 2 x + ( y + x d y d x ) + 2 y d y d x = 0 2 x + ( y + x d x d y ) + 2 y d x d y = 0 d y d x d x d y ( x + 2 y ) d y d x = − ( 2 x + y ) ( x + 2 y ) d x d y = − ( 2 x + y ) d y d x = − 2 x + y x + 2 y d x d y = − x + 2 y 2 x + y
문제 4. (표준) y 3 = x 2 y 3 = x 2 d y d x d x d y 답. 2 x 3 y 2 3 y 2 2 x 풀이. 양변을 미분하면 3 y 2 d y d x = 2 x 3 y 2 d x d y = 2 x d y d x = 2 x 3 y 2 d x d y = 3 y 2 2 x
음함수 미분으로 얻은 d y d x d x d y x , y x , y
예를 들어 원 x 2 + y 2 = 25 x 2 + y 2 = 2 5 ( 3 , 4 ) ( 3 , 4 ) d y d x = − x y d x d y = − y x − 3 4 − 4 3
y − 4 = − 3 4 ( x − 3 ) y − 4 = − 4 3 ( x − 3 )
입니다. 정리하면 3 x + 4 y = 25 3 x + 4 y = 2 5
문제 1. (표준) x 2 + y 2 = 10 x 2 + y 2 = 1 0 ( 1 , 3 ) ( 1 , 3 ) 답. − 1 3 − 3 1 풀이. d y d x = − x y d x d y = − y x ( 1 , 3 ) ( 1 , 3 ) − 1 3 − 3 1
문제 2. (표준) 곡선 x y = 2 x y = 2 ( 1 , 2 ) ( 1 , 2 ) 답. y = − 2 x + 4 y = − 2 x + 4 풀이. d y d x = − y x d x d y = − x y ( 1 , 2 ) ( 1 , 2 ) − 2 − 2 y − 2 = − 2 ( x − 1 ) y − 2 = − 2 ( x − 1 ) y = − 2 x + 4 y = − 2 x + 4
문제 3. (심화) 곡선 x 2 − x y + y 2 = 3 x 2 − x y + y 2 = 3 ( 1 , − 1 ) ( 1 , − 1 ) 답. 1 1 풀이. 양변을 미분하면 2 x − ( y + x d y d x ) + 2 y d y d x = 0 2 x − ( y + x d x d y ) + 2 y d x d y = 0 ( 2 y − x ) d y d x = y − 2 x ( 2 y − x ) d x d y = y − 2 x d y d x = y − 2 x 2 y − x d x d y = 2 y − x y − 2 x ( 1 , − 1 ) ( 1 , − 1 ) − 1 − 2 − 2 − 1 = − 3 − 3 = 1 − 2 − 1 − 1 − 2 = − 3 − 3 = 1
문제 4. (심화) 곡선 x 3 + y 3 = 9 x 3 + y 3 = 9 ( 1 , 2 ) ( 1 , 2 ) 답. − 1 4 − 4 1 풀이. 양변을 미분하면 3 x 2 + 3 y 2 d y d x = 0 3 x 2 + 3 y 2 d x d y = 0 d y d x = − x 2 y 2 d x d y = − y 2 x 2 ( 1 , 2 ) ( 1 , 2 ) − 1 4 − 4 1
f f f − 1 f − 1 f ( f − 1 ( x ) ) = x f ( f − 1 ( x ) ) = x x x
f ′ ( f − 1 ( x ) ) ⋅ ( f − 1 ) ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) ⋅ ( f − 1 ) ′ ( x ) = 1
이므로 다음이 나옵니다.
( f − 1 ) ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) ( f − 1 ) ′ ( x ) = f ′ ( f − 1 ( x ) ) 1
점으로 쓰면, f ( a ) = b f ( a ) = b ( f − 1 ) ′ ( b ) = 1 f ′ ( a ) ( f − 1 ) ′ ( b ) = f ′ ( a ) 1
직관은 이렇습니다. 역함수의 그래프는 원래 그래프를 y = x y = x ( f − 1 ) ′ ( b ) ( f − 1 ) ′ ( b ) f ′ f ′ b b a = f − 1 ( b ) a = f − 1 ( b )
예를 들어 f ( x ) = x 3 + x + 1 f ( x ) = x 3 + x + 1 ( f − 1 ) ′ ( 1 ) ( f − 1 ) ′ ( 1 ) f ( a ) = 1 f ( a ) = 1 a a a 3 + a + 1 = 1 a 3 + a + 1 = 1 a = 0 a = 0 f ′ ( x ) = 3 x 2 + 1 f ′ ( x ) = 3 x 2 + 1 f ′ ( 0 ) = 1 f ′ ( 0 ) = 1 ( f − 1 ) ′ ( 1 ) = 1 1 = 1 ( f − 1 ) ′ ( 1 ) = 1 1 = 1
문제 1. (기초) f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 ( f − 1 ) ′ ( x ) ( f − 1 ) ′ ( x ) 답. 1 2 2 1 풀이. f ′ ( x ) = 2 f ′ ( x ) = 2 ( f − 1 ) ′ ( x ) = 1 2 ( f − 1 ) ′ ( x ) = 2 1 f − 1 ( x ) = x − 3 2 f − 1 ( x ) = 2 x − 3 1 2 2 1
문제 2. (표준) f ( x ) = x 3 f ( x ) = x 3 ( f − 1 ) ′ ( 8 ) ( f − 1 ) ′ ( 8 ) 답. 1 12 1 2 1 풀이. f ( a ) = 8 f ( a ) = 8 a a a 3 = 8 a 3 = 8 a = 2 a = 2 f ′ ( x ) = 3 x 2 f ′ ( x ) = 3 x 2 f ′ ( 2 ) = 12 f ′ ( 2 ) = 1 2 ( f − 1 ) ′ ( 8 ) = 1 12 ( f − 1 ) ′ ( 8 ) = 1 2 1
문제 3. (표준) f ( x ) = x 2 + 1 f ( x ) = x 2 + 1 x ≥ 0 x ≥ 0 ( f − 1 ) ′ ( 5 ) ( f − 1 ) ′ ( 5 ) 답. 1 4 4 1 풀이. f ( a ) = 5 f ( a ) = 5 a ≥ 0 a ≥ 0 a 2 + 1 = 5 a 2 + 1 = 5 a = 2 a = 2 f ′ ( x ) = 2 x f ′ ( x ) = 2 x f ′ ( 2 ) = 4 f ′ ( 2 ) = 4 ( f − 1 ) ′ ( 5 ) = 1 4 ( f − 1 ) ′ ( 5 ) = 4 1
문제 4. (심화) f ( x ) = x + e x f ( x ) = x + e x ( f − 1 ) ′ ( 1 ) ( f − 1 ) ′ ( 1 ) 답. 1 2 2 1 풀이. f ( a ) = 1 f ( a ) = 1 a a a + e a = 1 a + e a = 1 a = 0 a = 0 0 + 1 = 1 0 + 1 = 1 a = 0 a = 0 f ′ ( x ) = 1 + e x f ′ ( x ) = 1 + e x f ′ ( 0 ) = 1 + 1 = 2 f ′ ( 0 ) = 1 + 1 = 2 ( f − 1 ) ′ ( 1 ) = 1 2 ( f − 1 ) ′ ( 1 ) = 2 1
곱·몫·거듭제곱이 복잡하게 얽힌 함수는 양변에 자연로그를 취한 뒤 미분하면 훨씬 간단해집니다. 로그가 곱을 합으로, 몫을 차로, 거듭제곱을 곱으로 바꾸기 때문입니다. 이것을 로그미분이라 합니다.
절차: (1) 양변에 ln ln x x y ′ y y y ′ y ′ = y × ( ⋯ ) y ′ = y × ( ⋯ )
예를 들어 y = x 2 x + 1 ( x − 1 ) 3 y = ( x − 1 ) 3 x 2 x + 1
ln y = 2 ln x + 1 2 ln ( x + 1 ) − 3 ln ( x − 1 ) ln y = 2 ln x + 2 1 ln ( x + 1 ) − 3 ln ( x − 1 )
미분하면
y ′ y = 2 x + 1 2 ( x + 1 ) − 3 x − 1 y y ′ = x 2 + 2 ( x + 1 ) 1 − x − 1 3
이므로 y ′ = y [ 2 x + 1 2 ( x + 1 ) − 3 x − 1 ] y ′ = y [ x 2 + 2 ( x + 1 ) 1 − x − 1 3 ]
문제 1. (표준) y = x 2 ( x + 1 ) 3 y = x 2 ( x + 1 ) 3 답. x ( x + 1 ) 2 ( 5 x + 2 ) x ( x + 1 ) 2 ( 5 x + 2 ) 풀이. ln y = 2 ln x + 3 ln ( x + 1 ) ln y = 2 ln x + 3 ln ( x + 1 ) y ′ y = 2 x + 3 x + 1 y y ′ = x 2 + x + 1 3 y ′ = x 2 ( x + 1 ) 3 [ 2 x + 3 x + 1 ] = 2 x ( x + 1 ) 3 + 3 x 2 ( x + 1 ) 2 y ′ = x 2 ( x + 1 ) 3 [ x 2 + x + 1 3 ] = 2 x ( x + 1 ) 3 + 3 x 2 ( x + 1 ) 2 x ( x + 1 ) 2 x ( x + 1 ) 2 x ( x + 1 ) 2 [ 2 ( x + 1 ) + 3 x ] = x ( x + 1 ) 2 ( 5 x + 2 ) x ( x + 1 ) 2 [ 2 ( x + 1 ) + 3 x ] = x ( x + 1 ) 2 ( 5 x + 2 )
문제 2. (표준) y = ( x + 1 ) 2 x 3 y = x 3 ( x + 1 ) 2 답. ( x + 1 ) 2 x 3 [ 2 x + 1 − 3 x ] x 3 ( x + 1 ) 2 [ x + 1 2 − x 3 ] 풀이. ln y = 2 ln ( x + 1 ) − 3 ln x ln y = 2 ln ( x + 1 ) − 3 ln x y ′ y = 2 x + 1 − 3 x y y ′ = x + 1 2 − x 3 y ′ = ( x + 1 ) 2 x 3 [ 2 x + 1 − 3 x ] y ′ = x 3 ( x + 1 ) 2 [ x + 1 2 − x 3 ]
문제 3. (심화) y = ( 2 x + 1 ) x y = ( 2 x + 1 ) x 2 x + 1 > 0 2 x + 1 > 0 답. ( 2 x + 1 ) x [ ln ( 2 x + 1 ) + 2 x 2 x + 1 ] ( 2 x + 1 ) x [ ln ( 2 x + 1 ) + 2 x + 1 2 x ] 풀이. ln y = x ln ( 2 x + 1 ) ln y = x ln ( 2 x + 1 ) y ′ y = ln ( 2 x + 1 ) + x ⋅ 2 2 x + 1 y y ′ = ln ( 2 x + 1 ) + x ⋅ 2 x + 1 2 y ′ = ( 2 x + 1 ) x [ ln ( 2 x + 1 ) + 2 x 2 x + 1 ] y ′ = ( 2 x + 1 ) x [ ln ( 2 x + 1 ) + 2 x + 1 2 x ]
음함수: 양변을 x x y y d y d x d x d y d y d x d x d y
d d x y n = n y n − 1 d y d x d x d y n = n y n − 1 d x d y d d x ( x y ) = y + x d y d x d x d ( x y ) = y + x d x d y 음함수 접선: d y d x d x d y ( x 0 , y 0 ) ( x 0 , y 0 )
역함수: ( f − 1 ) ′ ( b ) = 1 f ′ ( a ) ( f − 1 ) ′ ( b ) = f ′ ( a ) 1 a = f − 1 ( b ) a = f − 1 ( b )
역함수 문제는 먼저 f ( a ) = b f ( a ) = b a a
로그미분: 곱·몫·거듭제곱은 ln ln y y
문제 1. (기초) x 2 − y 2 = 9 x 2 − y 2 = 9 d y d x d x d y 답. x y y x 풀이. 양변을 미분하면 2 x − 2 y d y d x = 0 2 x − 2 y d x d y = 0 d y d x = x y d x d y = y x
문제 2. (기초) f ( x ) = 5 x − 2 f ( x ) = 5 x − 2 ( f − 1 ) ′ ( 3 ) ( f − 1 ) ′ ( 3 ) 답. 1 5 5 1 풀이. f ′ ( x ) = 5 f ′ ( x ) = 5 ( f − 1 ) ′ ( 3 ) = 1 5 ( f − 1 ) ′ ( 3 ) = 5 1
문제 3. (표준) x 3 + y 3 = 6 x y x 3 + y 3 = 6 x y d y d x d x d y 답. 2 y − x 2 y 2 − 2 x y 2 − 2 x 2 y − x 2 풀이. 양변을 미분하면 3 x 2 + 3 y 2 d y d x = 6 ( y + x d y d x ) 3 x 2 + 3 y 2 d x d y = 6 ( y + x d x d y ) d y d x d x d y ( 3 y 2 − 6 x ) d y d x = 6 y − 3 x 2 ( 3 y 2 − 6 x ) d x d y = 6 y − 3 x 2 d y d x = 6 y − 3 x 2 3 y 2 − 6 x = 2 y − x 2 y 2 − 2 x d x d y = 3 y 2 − 6 x 6 y − 3 x 2 = y 2 − 2 x 2 y − x 2
문제 4. (표준) sin y + x = y sin y + x = y d y d x d x d y 답. 1 1 − cos y 1 − cos y 1 풀이. 양변을 미분하면 cos y d y d x + 1 = d y d x cos y d x d y + 1 = d x d y d y d x d x d y 1 = ( 1 − cos y ) d y d x 1 = ( 1 − cos y ) d x d y d y d x = 1 1 − cos y d x d y = 1 − cos y 1
문제 5. (표준) e y = x + y e y = x + y d y d x d x d y 답. 1 e y − 1 e y − 1 1 풀이. 양변을 미분하면 e y d y d x = 1 + d y d x e y d x d y = 1 + d x d y ( e y − 1 ) d y d x = 1 ( e y − 1 ) d x d y = 1 d y d x = 1 e y − 1 d x d y = e y − 1 1
문제 6. (표준) f ( x ) = x 3 + 2 x + 1 f ( x ) = x 3 + 2 x + 1 ( f − 1 ) ′ ( 4 ) ( f − 1 ) ′ ( 4 ) 답. 1 5 5 1 풀이. f ( a ) = 4 f ( a ) = 4 a a a 3 + 2 a + 1 = 4 a 3 + 2 a + 1 = 4 a 3 + 2 a − 3 = 0 a 3 + 2 a − 3 = 0 a = 1 a = 1 1 + 2 − 3 = 0 1 + 2 − 3 = 0 f ′ ( x ) = 3 x 2 + 2 f ′ ( x ) = 3 x 2 + 2 f ′ ( 1 ) = 5 f ′ ( 1 ) = 5 ( f − 1 ) ′ ( 4 ) = 1 5 ( f − 1 ) ′ ( 4 ) = 5 1
문제 7. (표준) 곡선 x 2 + 4 y 2 = 8 x 2 + 4 y 2 = 8 ( 2 , 1 ) ( 2 , 1 ) 답. y = − 1 2 x + 2 y = − 2 1 x + 2 풀이. 양변을 미분하면 2 x + 8 y d y d x = 0 2 x + 8 y d x d y = 0 d y d x = − x 4 y d x d y = − 4 y x ( 2 , 1 ) ( 2 , 1 ) − 2 4 = − 1 2 − 4 2 = − 2 1 y − 1 = − 1 2 ( x − 2 ) y − 1 = − 2 1 ( x − 2 ) y = − 1 2 x + 2 y = − 2 1 x + 2
문제 8. (심화) 곡선 x 2 y + x y 2 = 6 x 2 y + x y 2 = 6 ( 1 , 2 ) ( 1 , 2 ) 답. − 8 5 − 5 8 풀이. 양변을 미분합니다. d d x ( x 2 y ) = 2 x y + x 2 d y d x d x d ( x 2 y ) = 2 x y + x 2 d x d y d d x ( x y 2 ) = y 2 + x ⋅ 2 y d y d x d x d ( x y 2 ) = y 2 + x ⋅ 2 y d x d y 2 x y + y 2 + ( x 2 + 2 x y ) d y d x = 0 2 x y + y 2 + ( x 2 + 2 x y ) d x d y = 0 ( 1 , 2 ) ( 1 , 2 ) 2 ⋅ 1 ⋅ 2 + 4 + ( 1 + 4 ) d y d x = 0 2 ⋅ 1 ⋅ 2 + 4 + ( 1 + 4 ) d x d y = 0 8 + 5 d y d x = 0 8 + 5 d x d y = 0 d y d x = − 8 5 d x d y = − 5 8
문제 9. (심화) y = x ln x y = x l n x x > 0 x > 0 답. x ln x ⋅ 2 ln x x x l n x ⋅ x 2 ln x 풀이. 양변에 로그를 취하면 ln y = ln x ⋅ ln x = ( ln x ) 2 ln y = ln x ⋅ ln x = ( ln x ) 2 y ′ y = 2 ln x ⋅ 1 x = 2 ln x x y y ′ = 2 ln x ⋅ x 1 = x 2 ln x y ′ = x ln x ⋅ 2 ln x x y ′ = x l n x ⋅ x 2 ln x
문제 10. (심화) x + y = 4 x + y = 4 ( 4 , 4 ) ( 4 , 4 ) d y d x d x d y 답. − 1 − 1 풀이. 양변을 미분하면 1 2 x + 1 2 y d y d x = 0 2 x 1 + 2 y 1 d x d y = 0 d y d x = − y x d x d y = − x y ( 4 , 4 ) ( 4 , 4 ) − 2 2 = − 1 − 2 2 = − 1
문제 11. (심화) y = x − 1 ( x + 2 ) 3 ( x + 1 ) 4 y = ( x + 1 ) 4 x − 1 ( x + 2 ) 3 y y 답. y [ 1 2 ( x − 1 ) + 3 x + 2 − 4 x + 1 ] y [ 2 ( x − 1 ) 1 + x + 2 3 − x + 1 4 ] 풀이. ln y = 1 2 ln ( x − 1 ) + 3 ln ( x + 2 ) − 4 ln ( x + 1 ) ln y = 2 1 ln ( x − 1 ) + 3 ln ( x + 2 ) − 4 ln ( x + 1 ) y ′ y = 1 2 ( x − 1 ) + 3 x + 2 − 4 x + 1 y y ′ = 2 ( x − 1 ) 1 + x + 2 3 − x + 1 4 y ′ = y [ 1 2 ( x − 1 ) + 3 x + 2 − 4 x + 1 ] y ′ = y [ 2 ( x − 1 ) 1 + x + 2 3 − x + 1 4 ]
문제 12. (심화) f ( x ) = tan x f ( x ) = tan x − π 2 < x < π 2 − 2 π < x < 2 π ( f − 1 ) ′ ( 1 ) ( f − 1 ) ′ ( 1 ) 답. 1 2 2 1 풀이. f ( a ) = 1 f ( a ) = 1 a a tan a = 1 tan a = 1 a = π 4 a = 4 π f ′ ( x ) = 2 x = 1 2 x f ′ ( x ) = sec 2 x = cos 2 x 1 cos π 4 = 1 2 cos 4 π = 2 1 f ′ ( π 4 ) = 1 ( 1 / 2 ) 2 = 2 f ′ ( 4 π ) = ( 1 / 2 ) 2 1 = 2 ( f − 1 ) ′ ( 1 ) = 1 2 ( f − 1 ) ′ ( 1 ) = 2 1 1 1 + x 2 1 + x 2 1 x = 1 x = 1
y y d y d x d x d y x 2 + y 2 = r 2 x 2 + y 2 = r 2 d y d x = − x y d x d y = − y x 음함수로 주어진 곡선의 접선을 점을 대입해 구할 수 있는가?
역함수 도함수 공식 ( f − 1 ) ′ ( b ) = 1 f ′ ( a ) ( f − 1 ) ′ ( b ) = f ′ ( a ) 1 a a
f ( x ) = x 3 f ( x ) = x 3 ( f − 1 ) ′ ( 8 ) ( f − 1 ) ′ ( 8 ) 1 12 1 2 1 로그미분의 4단계 절차를 말할 수 있는가?
y = x x y = x x y = ( 2 x + 1 ) x y = ( 2 x + 1 ) x x x x y = 1 x y = 1 d y d x d x d y y = 1 / x y = 1 / x − 1 x 2 − x 2 1