(2000 )
|
|
|
- いおり みつだ
- 4 years ago
- Views:
Transcription
1 (000)
2 < > = = = (BC 67» BC 1) (= ) 18 ( 00 ) ( ¼"½ '"½ &) ¼ 18 ¼ 0 ¼ =3: ¼ 1 5cm ` ¼ = ` 5 = ` 10 () ` =10¼ (cm) (1) 3cm () r () () (1) r () r 1 4 (3) r, 60 ± 1
3 < > µ AB ` µ ± = ` (1) µ =360 ± ¼ 360 ± =¼ ¼ ; 3:14 () µ =180 ± ¼ 180 ± = ¼ (3) µ =90 ± 1 4 ¼ 90 ± = ¼ µ = 180± ¼ ; 57:3± 360 ± ; 180 ± ; 90 ±
4 < > 1 1 ( ) 0 ± 360 ± (1) 40 ± =360 ± +60 ± =¼ + ¼ 3 () =7 3 ¼ () () 510 ± = 360 ± 150 ± = ¼ 5 6 ¼ () = 17 6 ¼ () (1) 540 ± () 40 ± (3) 600 ± (4) 390 ± (5) 840 ± (6) 405 ± 3 1 () (1) r ` () r S ` = S = 3
5 < 3> µ r OAB ` OAB S (1) µ =¼= 360 ± ` ` =¼r S S = ¼r () µ = ¼= 180 ± (1) ` = ¼r S = 1 ¼r (3) µ = ¼ = 90± (1) 1 4 ` = 1 ¼r S = 1 4 ¼r µ ` OAB S µ r ` = S = 4
6 < > y =sinx y =cosx y =tanx (1) y =sinx () y =cosx (3) y =tanx 5
7 < > A; B; C y = A sin(bx + C) ³ sin x + ¼ =sinxcos ¼ +cosxsin ¼ cos ¼ =cos90± =0; sin ¼ =sin90± =1 ³ sin x + ¼ =cosx y =cosx y =sinx y =cosx y =cosx y =sinx x ¼ cos x sin x ¼ sin x cos x ¼ y = A sin(bx + C) Bx + C ³ y =sin x ¼ 6
8 < > y =sinx y =sinx y =sinx y =sinx y =sinx y x y = 3sinx 7
9 < > y =sinx ¼ x y =sinx ¼ y =sin(x) ¼ y =sin(x) ¼ y =sin(3x) 8
10 < > p y = p x p x 0 x y p x ( x = 0) p 0 y y = 0 y y = p x +1 p 0 x +1 = 0 ) x = 1 : x = 1 p = 0 : y = 0 (1) y = p x + () y = p x 1 9
11 < > y = p 3 x p 0 3 x = 0 ) 3 = x : x 5 3 p = 0 : y = 0 y = p x 1 p 0 x 1 = 0 ) x = 1 : x = 1 p = 0 p 5 0 : y 5 0 (1) y = p x +1 () y = p x + 10
12 < 3> y = p x +1 x +1= 0 : x = 1 p p 5 = 0 : y 5 y = 1+ p x x = 0 := x p = 0 1+ p = 1 : y = 1 (1) y = 1+ p x +1 () y =1 p x 11
13 < > y = 1 x x =0 0 0 : x 6= 0 y = 1 x xy =1 y 0 y = 1 x 6=0 y 0 : y 6= 0 y = 1 x 0 x 6= 0 : x 6= : y 6= 0 y = 1 x x + y = 1 x +1 1
14 < > y =3+ 1 x 6= 0 : x 6= 1 6=0 y 3= 1 x 6=0 y 3 6= 0 : y 6= 3 y = 1 x x +y +3 x x =x y =3 x =;y =3 y =3+ 1 x y =+ 1 x +1 13
15 < 1> y x y = f(x) x y x y y = f(x) x y f(x) =+ p x 1 p 0 x 1 = 0 f(x) x = 1 y =+ p x 1 p = 0 y = f(x) =+ 1 x 1 0 x 1 6= 0 x 6= 1 f(x) 1 y =+ 1 x 1 1 6=0 y 6= x 1 (1) f(x) = p x 1 () f(x) =+ p 1 x (3) f(x) =1+ 1 x + 14
16 < > f(x) =log 4 (x ) =log 4 =4 > 0 = x > 0 f(x) x> y =log 4 (x ) x =4 y y f(x) = x +1 x f(x) y = x +1 x > 0 x +1> 1 y>1 (1) f(x) =log (1 x) () f(x) = x 1 (3) f(x) = 3 x 15
17 < 3> 1 f(x) =3+sinx x f(x) 1 5 sin x sinx sinx sinx y 5 5 f(x) =1+3cos(x) cos x x f(x) 1 5 cos(x) cos(x) cos(x) cos(x) y 5 4 (1) f(x) =3+sinx () f(x) =1+cosx (3) f(x) =+3sinx (4) f(x) =+4cos(3x) 16
18 < 4> 1 µ tan µ = Y X µ 90 ± 90 ± x 0(X =0) 0 tan µ 90 ± ; 70 ± ; 450 ± ; tan µ µ 6= ¼ n¼ (n =0; 1; ; 3; ) y =tanx 5 5 f(x) =tan(x) tan( ) 6= ¼ n¼ (n =0; 1; ; ) x 6= ¼ n¼ x 6= ¼ 4 n ¼ f(x) ¼ 4 n ¼(n ) (1) f(x) =tan(3x) () f(x) =tan(¼x) 17
19 < > y = f(x) x y ( ) x 1 6= x f(x 1 ) 6= f(x ) y = f(x) 1 1 f(x) =x 1 y = f(x) 1 1 f(x) =x y = f(x) 1 1 x 1 = 1 ;x =1 f(x 1 )=f(x )=1 ( ) x 1 ;x (1) y =3x () y = x 3 x (3) y = 1 x (x 6= 0) 18
20 < > f(x) 1 1 y b b = f(a) x a 1 a = f 1 (b) f(x) =x 1 y = f(x) 1 1 b = f(a) f(a) =a 1 b =a 1 a a = 1 b + 1 a = f 1 (b) f 1 (b) = 1 b + 1 f(x) y = f(x) 1 1 f 1 (b) b (1) f(x) =3x () f(x) = 1 x + (3)f(x) =p x 19
21 < > y = f(x) 1 1 y b x f 1 (b) y = f 1 (x) y = f(x) f(x) =x +1 b = f(a) () a = f 1 (b) b =a +1 () a = 1 b 1 = f 1 (b) f 1 (x) = 1 x 1 y = f(x) y = f 1 (x) y = x f(x) f 1 (x) (1) f(x) =3x + ()f(x) = 1 x (3) f(x) = p x 0
22 < > y = f(x) 1 1 f f 1 f f 1 f(x) =x +1f x = 0 y = f(x) 1 1 b = f(a) () a = f 1 (b) b = a +1 () a = p b 1=f 1 (b) (a = 0 ;b= 1) (a = 0 ;b= 1) b x f 1 (x) = p x 1 x = 1 y = f(x) y = f 1 (x) y = x f(x) =(x +1) f x = 1 y = f(x) 1 1 1
23 < > y = f(x) 1 1 y = f(x) (a ; b) b = f(a) () a = f 1 (b) (b ; a) y = f 1 (x) y = f 1 (x) y = f(x) y = x f(x) =3 x b = f(a) () a = f 1 (b) b =3 a () a =log 3 b = f 1 (b) f 1 (x) =log 3 x y =log 3 x y =3 x y =log 3 x y =3 x y = x µ 1 x y = y =log1 x
24 < > y =sinx 1 5 y 5 1 ¼ 5 x 5 ¼ 1 1 y =sinx ¼ 5 x 5 ¼ ; 1 5 y 5 1 y =sin 1 x y =arcsinx 1 5 x 5 1 ; ¼ 5 y 5 ¼ y =sin 1 x y =sinx y = x y =sin 1 x a =sin 1 b () b =sina µ 1 sin 1 µ µ 1 µ =sin 1 () 1 =sinµ ¼ 5 µ 5 ¼ sin 1 µ µ = ¼ 6 sin 1 µ 1 = ¼ 6 Ãp! (1) sin 1 = () sin 1 Ãp 3! µ = (3) sin 1 1 = 3
25 < > y =cosx 1 5 y x 5 ¼ 1 1 y =cosx 0 5 x 5 ¼; 1 5 y 5 1 y =cos 1 x y =arccosx 1 5 x 5 1 ; 0 5 y 5 ¼ y =cos 1 x y =cosx y = x y =cos 1 x a =cos 1 b () b =cosa µ 1 cos 1 µ µ 1 µ =cos 1 () 1 =cosµ 0 5 µ 5 ¼ cos µ 1 µ µ = ¼ 3 cos 1 µ 1 = ¼ 3 Ãp! Ã p! µ 3 (1) cos 1 = () cos 1 = (3) cos 1 1 = 4
26 < > y =tanx ¼ + n¼n ¼ <x<¼ 1 1 y =tanx : ¼ <x<¼ ; : y =tan 1 x y = arctan x : ; : ¼ <y<¼ y =tan 1 x y =tanx y = x tan 1 x a =tan 1 b () b =tana tan 1 ( p 3) µ µ =tan 1 ( p 3) () p 3=tanµ ¼ <µ<¼ tan µ p 3 µ µ = ¼ 3 tan 1 ( p 3) = ¼ 3 (1) tan 1 (1) = () tan 1 Ãp 3 3! = (3) tan 1 p 3 = 5
27 < > a 1 ;a ;a 3 ; ;a n a n n fa n g fa n g n a n n n!1 1 1 ; 3 ; 4 3 ; ; n +1 ; n n!1a n = n +1 n 1 1 ; 1 ; 1 4 ; ; µ 1 n 1 ; µ n!1a n = 1 n ; 3 4 ; 5 6 ; ; n 1 n ; 6
28 < > fa n g n!1a n fa n g n!1 a n! lim a n = n!1 fa n g lim n!1 n +1 n µ =1; lim 1 n 1 =0 n!1 n!1 1 n! 0 µ n +1 lim = lim 1+ 1 =1; n!1 n n!1 n lim n!1 3n n +1 = lim n! n = 3 +0 = 3 n!1 n! 0; 1 n! 0 n +n 1+ lim n!1 3n +1 = lim n n! = = 1 3 n. (1) n +1 lim n!1 n 1 = () lim n!1 n +3n +1 n +1 = 7
29 < > fa n g fa n g 1; 4; 9; 16; ;n ; n!1a n = n fa n g n!1 a n!1 ( a n! +1) lim a n = 1 ( lim a n =+1) n!1 n!1 ; 0; ; ; 4 n; n!1a n =4 n fa n g n!1 a n! 1 lim a n = 1 n!1 lim n!1 4 n = 1 (1) lim n!1 (n 3 + n) = () lim n!1 (5 n ) = 8
30 < 4> 1 1; 1; 1; 1; ; ( 1) n ; 1; ; 4; 8; ; ( ) n 1 ; n fa n g (1) lim a n =+1 n!1 () lim a n = 1 n!1 (3) (1) lim n!1 µ 4 n µ =+1 () lim 3 n = 0 (3) lim 3 n!1 4 n!1 (4) 4 3 ; 16 µ 9 ; ; 4 n ; 3 µ 4 n = 1 3 (1) 3 ; 9 4 ; 7 8 ; 81 µ n 3 16 ; ; ; () 3 ; 4 9 ; 8 7 ; 16 µ n 81 ; ; ; 3 (3) 3 ; 4 9 ; 8 7 ; 16 µ 81 ; ; 3 n ; (4) 3 ; 9 4 ; 7 8 ; 81 µ 16 ; ; 3 n ; (5) 3 µ n 3 ; 9 4 ; 7 8 ; ; ; ; 9
31 < > fa n g (1) a 1 + a + a a n + fa n g S n = a 1 + a + a a n n S 1 ;S ;S 3 ; ;S n ; S ( lim S n =) n!1 (1) S a 1 + a + a a n + = S S n + S n S n = n 1 n 1 S n = 1 S n = n n+1 S n =1 1 n n! 1 1 n! 0 1 n+1 S = lim S n = lim µ1 1 =1 n!1 n!1 n () S n + 30
32 < > =0: =0: () 0:333 =0:_3 ; 0: =0:_3_6 0: = 0:5_1_3 0:_1_ =S S = 0:_1_ =0:111 = 0:1 + 0: : µ µ 1 1 µ 1 3 = n S n µ µ µ n 1 µ S n = µ 1 1 µ 1 3 µ 1 n µ S n = µ µ n S n = S n = µ 1 n µ 1 n n!1! ½ 1 S = lim S n = lim n!1 n! µ 1 n ¾ 99 = = 4 33 n n+1 S =0:_7 =0:777 31
33 < > f(x) x a a f(x) x! a f(x)! lim f(x) = x!a x! a f(x) a x 1 lim x 3x 3x =10; lim x!5 x!1 x +1 = 3 ; lim h!0 h +3 h +1 =3 3 x + x (x 1)(x +) lim = lim x!1 x 1 x!1 (x 1)(x +1) = lim x + x!1 x +1 = 3 (3 + h) 3 lim h!0 h =lim h!0 (9 + 6h + h ) 9 h = lim h!0 (6 + h) =6 7x +3 (1) lim x! x +1 = x x () lim = x! x 4 x x 3 (3) lim = x!3 x 3 (4) lim h!0 (4 + h) 4 h = 3
34 < > (a + b) 3 =(a + b)(a + b) =(a + b)(a +ab + b ) = a(a +ab + b )+b(a +ab + b ) = a 3 +a b + ab + ba +ab + b 3 = a 3 +3a b +3ab + b 3 (1) (a + b) 4 =(a + b)(a + b) 3 =(a + b)(a 3 +3a b +3ab + b 3 ) = a 4 + a 3 b + a b + ab 3 + b 4 ³ () (a + b) 5 =(a + b) a 4 + a 3 b + a b + ab 3 + b 4 = a 5 + a 4 b + a 3 b + a b 3 + ab 4 + b 5 (a + b) n (a + b) 0 =1 1 (a + b) 1 =1 a +1 b 1 1 (a + b) =1 a + ab +1 b 1 1 (a + b) 3 =1 a 3 +3 a b +3 ab +1 b (a + b) 4 = a 4 + a 3 b + a b + ab 3 + b 4 (a + b) 5 = a 5 + a 4 b + a 3 b + a b 3 + ab 4 + b 5 (a + b) 6 (a + b) 6 = a 6 + a 5 b + a 4 b + a 3 b 3 + a b 4 + ab 5 + b 6 33
35 < > f(x) x f(x) f(x) f 0 (x) f 0 (x) =lim h!0 f(x + h) f(x) h f(x) 1 f(x) =1 (1) 0 = f 0 f(x + h) f(x) 1 1 (x) = lim =lim =0 h!0 h h!0 h f(x) =x (x) 0 = f 0 f(x + h) f(x) x + h x h (x) =lim = lim = lim h!0 h h!0 h h!0 h =1 3 f(x) =x x 0 = f 0 f(x + h) f(x) (x + h) x (x) =lim = lim h!0 h h!0 h =lim h!0 x +xh + h x h = lim h!0 (x + h) =x 4 f(x) =x 3 x 3 0 = f 0 f(x + h) f(x) (x + h) 3 x 3 (x) =lim = lim h!0 h h!0 h =lim h!0 x 3 +3x h +3xh + h 3 x 3 h =lim h!0 (3x +3xh + h )=3x f(x) =x 4 f(x) (x 4 ) 0 = f 0 (x) = 34
36 < > f(x) =x 5 f 0 (x) () (x 5 ) 0 = f 0 (x) = f(x) =x 6 f 0 (x) () (x 6 ) 0 = f 0 (x) = 3 x 0 =1 f(x) x 0 x 1 x x 3 x 4 x 5 x 6 f 0 (x) 4 n x n (x n ) 0 (x n ) 0 = 35
37 < 3 > f(x);g(x) k kf(x) 0 = k f 0 (x) f(x)+g(x) 0 = f 0 (x)+g 0 (x) f(x) g(x) 0 = f 0 (x) g 0 (x) () () () (1) (x 5 + x 6 ) 0 =(x 5 ) 0 +(x 6 ) 0 =5x 4 +6x 5 () (7x 4 ) 0 =7 (x 4 ) 0 =7 4x 3 =8x 3 (3) (6x 5 +5x 4 ) 0 =(6x 5 ) 0 +(5x 4 ) 0 =30x 4 +0x 3 (4) (x 7 4x 5 +5x 8) 0 =(x 7 ) 0 (4x 5 ) 0 +(5x ) 0 (8) 0 =7x 6 0x 4 +10x (5) (x +3)(x 4) 0 =(x 4 x 1) 0 =4x 3 x (1) (x 3 x 5 ) 0 () (1x 5 ) 0 (3) (4x 3 +8x 5 ) 0 (4) (7x 6 3x 4 +5) 0 (5) (10x 4 5x ) 0 (6) (4x 6 7x 4 +8x 3 9x) 0 (7) (x )(x +3) 0 (8) (x 4)(x 5) 0 36
38 < 1 > f(x) a x f(a) >f(x) f(x) x = a f(x) b x f(b) <f(x) f(x) x = b f(b) 3 y =x 3 9x +1x y 0 =6x 18x +1 =6(x 1)(x ) x = 1 x = y 0 =0 x 1 y y % 3 & % x =1 y =3 x = y = 3 y = x 3 3x x = y = x = y = x y 0 y 37
39 < > 4 y =3x 4 16x 3 +18x +8 3 y 0 =1x 3 48x +36x =1x(x 4x +3) =1x(x 1)(x 3) x =0; x =1; x =3 y 0 =0 x y y & 8 % 13 & 19 % x =1 y =13 x =0 y =8 x =3 y = 19 (1) y = x 4 x +3 x y 0 y () y = 3x 4 +4x 3 +1x x y 0 y 38
40 < > f(x) = 1 x +3 f 0 (x) f(x) = 1 x +3 f(x + h) = 1 x + h +3 µ 1 0 = f 0 f(x + h) f(x) (x) = lim = lim x +3 h!0 h h!0 1 x + h +3 1 x +3 h =lim h!0 (x +3) (x + h +3) (x + h +3)(x +3) h =lim h!0 h (x + h +3)(x +3) h =lim h!0 1 (x + h +3)(x +3) 1 =lim h!0 1 (x + h +3)(x +3) = 1 (x +0+3)(x +3) = 1 (x +3) µ = x +3 (x +3) ³ (x +3) 1 0 = (x +3) f(x) = 1 x (1) f 0 (x) ( ) f 0 (x) = () (1) 39
41 < > f(x) = p x +3 f 0 (x) f(x) = p x +3 f(x + h) = p x + h +3 ³ p p px +3 0 = f 0 f(x + h) f(x) x + h +3 x +3 (x) =lim = lim h!0 h h!0 h p p p p x + h +3 x +3 x + h +3+ x +3 = lim h!0 h p x + h +3+ p x +3 p p x + h +3 x +3 = lim h!0 h p x + h +3+ p x +3 (x + h +3) (x +3) = lim h!0 h p x + h +3+ p x +3 h = lim h!0 h p x + h +3+ p x +3 = lim 1 p p h!0 x + h +3+ x +3 = 1 p x p x +3 = 1 p x +3 ³ px +3 0 = 1 p x +3 ³ 0 (x +3) 1 = 1 (x +3) 1 f(x) = p x (1) f 0 (x) () f 0 (x) = () (1) 40
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)
< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3) 6 y = g(x) x = 1 g( 1) = 2 ( 1) 3 = 2 ; g 0 ( 1) =
No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π
4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan
, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
電気電子数学入門 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/073471 このサンプルページの内容は, 初版 1 刷発行当時のものです. i 14 (tool) [ ] IT ( ) PC (EXCEL) HP() 1 1 4 15 3 010 9 ii 1... 1 1.1 1 1.
no35.dvi
p.16 1 sin x, cos x, tan x a x a, a>0, a 1 log a x a III 2 II 2 III III [3, p.36] [6] 2 [3, p.16] sin x sin x lim =1 ( ) [3, p.42] x 0 x ( ) sin x e [3, p.42] III [3, p.42] 3 3.1 5 8 *1 [5, pp.48 49] sin
2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta
009 IA 5 I, 3, 4, 5, 6, 7 6 3. () Arcsin ( (4) Arccos ) 3 () Arcsin( ) (3) Arccos (5) Arctan (6) Arctan ( 3 ) 3. n () tan x (nπ π/, nπ + π/) f n (x) f n (x) fn (x) Arctan x () sin x [nπ π/, nπ +π/] g n
. p.1/15
. p./5 [ ] x y y x x y fx) y fx) x y. p.2/5 [ ] x y y x x y fx) y fx) x y y x y x y fx) f y) x f y) f f f f. p.2/5 [ ] x y y x x y fx) y fx) x y y x y x y fx) f y) x f y) f f f f [ ] a > y a x R ). p.2/5
A A p.1/16
A A p./6 A p.2/6 [ ] x y y x x y fx) y fx) x y A p.3/6 [ ] x y y x x y fx) y fx) x y y x y x y fx) f y) x f y) f f f f A p.3/6 [ ] x y y x x y fx) y fx) x y y x y x y fx) f y) x f y) f f f f [ ] a > y
t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ
1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ
1 (1) ( i ) 60 (ii) 75 (iii) 15 () ( i ) (ii) 4 (iii) 7 1 ( () r, AOB = θ 0 < θ < ) OAB A OB P ( AB ) < ( AP ) (4) 0 < θ < sin θ < θ < tan θ 0 x, 0 y (1) sin x = sin y (x, y) () cos x cos y (x, y) 1 c
f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a
3 3.1 3.1.1 A f(a + h) f(a) f(x) lim f(x) x = a h 0 h f(x) x = a f 0 (a) f 0 (a) = lim h!0 f(a + h) f(a) h = lim x!a f(x) f(a) x a a + h = x h = x a h 0 x a 3.1 f(x) = x x = 3 f 0 (3) f (3) = lim h 0 (
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
- II
- II- - -.................................................................................................... 3.3.............................................. 4 6...........................................
D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
http://know-star.com/ 3 1 7 1.1................................. 7 1.2................................ 8 1.3 x n.................................. 8 1.4 e x.................................. 10 1.5 sin
4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X
4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
高等学校学習指導要領解説 数学編
5 10 15 20 25 30 35 5 1 1 10 1 1 2 4 16 15 18 18 18 19 19 20 19 19 20 1 20 2 22 25 3 23 4 24 5 26 28 28 30 28 28 1 28 2 30 3 31 35 4 33 5 34 36 36 36 40 36 1 36 2 39 3 41 4 42 45 45 45 46 5 1 46 2 48 3
Acrobat Distiller, Job 128
(2 ) 2 < > ( ) f x (x, y) 2x 3+y f y (x, y) x 2y +2 f(3, 2) f x (3, 2) 5 f y (3, 2) L y 2 z 5x 5 ` x 3 z y 2 2 2 < > (2 ) f(, 2) 7 f x (x, y) 2x y f x (, 2),f y (x, y) x +4y,f y (, 2) 7 z (x ) + 7(y 2)
1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29
![[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29](/thumbs/93/114156928.jpg "[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29")
A p./29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x) + C f(x) A p.2/29 [ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx [ ] F(x) f(x) C F(x)
J1-a.dvi
4 [ ] 4. ( ) (x ) 3 (x +). f(x) (x ) 3 x 3 3x +3x, g(x) x + f(x) g(x) f(x) (x 3)(x +)+x+ (x 3)g(x)+x+ g(x) r(x) x+ g(x) x (x+)+ x r(x)+ g(x) x x 4x+5 r(x) g(x) x (f(x) (x 3)g(x)) g(x) x f(x) f(x)g(x) 4
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a
... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c
di-problem.dvi
005/05/05 by. I : : : : : : : : : : : : : : : : : : : : : : : : :. II : : : : : : : : : : : : : : : : : : : : : : : : : 3 3. III : : : : : : : : : : : : : : : : : : : : : : : : 4 4. : : : : : : : : : :
I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x
11 11.1 I y = a I a x I x = a + 1 f(a) x a = f(a +) f(a) (11.1) x a 0 f(a) f(a +) f(a) = x a x a 0 (11.) x = a a f (a) d df f(a) (a) I dx dx I I I f (x) d df dx dx (x) [a, b] x a ( 0) x a (a, b) () [a,
function2.pdf
2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)
18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C
8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,
DVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED)
rational number p, p, (q ) q ratio 3.14 = 3 + 1 10 + 4 100 ( ) a, b c a 2 + b 2 = c 2. 2 1 2 2 : 2 2 = p q, p, q 2q 2 = p 2. p 2 p 2 2 2 q 2 p, q (QED) ( a) ( b) a > b > 0 a < nb n A A B B A A, B B A =
(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)
2017 12 9 4 1 30 4 10 3 1 30 3 30 2 1 30 2 50 1 1 30 2 10 (1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10) (1) i 23 c 23 0 1 2 3 4 5 6 7 8 9 a b d e f g h i (2) 23 23 (3) 23 ( 23 ) 23 x 1 x 2 23 x
A S- hara/lectures/lectures-j.html r A = A 5 : 5 = max{ A, } A A A A B A, B A A A %
A S- http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html r A S- 3.4.5. 9 phone: 9-8-444, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office
S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d
S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....
17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
行列代数2010A
a ij i j 1) i +j i, j) ij ij 1 j a i1 a ij a i a 1 a j a ij 1) i +j 1,j 1,j +1 a i1,1 a i1,j 1 a i1,j +1 a i1, a i +1,1 a i +1.j 1 a i +1,j +1 a i +1, a 1 a,j 1 a,j +1 a, ij i j 1,j 1,j +1 ij 1) i +j a
2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p
2012 IA 8 I 1 10 10 29 1. [0, 1] n x = 1 (n = 1, 2, 3,...) 2 f(x) = n 0 [0, 1] 2. 1 x = 1 (n = 1, 2, 3,...) 2 f(x) = n 0 [0, 1] 1 0 f(x)dx 3. < b < c [, c] b [, c] 4. [, b] f(x) 1 f(x) 1 f(x) [, b] 5.
1 I
1 I 3 1 1.1 R x, y R x + y R x y R x, y, z, a, b R (1.1) (x + y) + z = x + (y + z) (1.2) x + y = y + x (1.3) 0 R : 0 + x = x x R (1.4) x R, 1 ( x) R : x + ( x) = 0 (1.5) (x y) z = x (y z) (1.6) x y =
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
06微分 [更新済み]
![06微分 [更新済み]](/thumbs/100/144040181.jpg "06微分 [更新済み]")
T Xclub E 微分 が よくわからないときに開く本 例題で式の計算がよくわかる! 改訂版 内容 初等関数の微分積 商の微分合成関数 逆関数の微分関数の増減 高知工科大学 KOCHI UNIVERSITY OF T ECHNOLOGY 井上昌昭著 Copyrigt(C) Masaaki Inoue < 関数の定義域と値域 > 例 無理関数 y = x の場合, の中が負になってはいけない という制限が自動的につく
1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +
6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +
1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1
sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V
DVIOUT-HYOU
() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.
04.dvi
22 I 4-4 ( ) 4, [,b] 4 [,b] R, x =, x n = b, x i < x i+ n + = {x,,x n } [,b], = mx{ x i+ x i } 2 [,b] = {x,,x n }, ξ = {ξ,,ξ n }, x i ξ i x i, [,b] f: S,ξ (f) S,ξ (f) = n i= f(ξ i )(x i x i ) 3 [,b] f:,
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F
III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ
ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y
01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y
70 : 20 : A B (20 ) (30 ) 50 1
70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,
6,,3,4,, 3 4 8 6 6................................. 6.................................. , 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p,
1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =
1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A
total2010.dvi
Ô ØÖ ÁÒØ ÖÔÓÐ Ø ÓÒ ÔÓÐÝÒÑ Ð Ø ÜØÖ ÔÓÐ Ø ÓÒ ËÓÑÑ Ö º½ ÁÒØ ÖÔÓÐ Ø ÓÒ Ä Ö Ò º º º º º º º º º º º º º º º º º º ¾ º½º½ ÓÖÑÙÐ Ø ÓÒ ÖÝ ÒØÖ ÕÙ º º º º º º º º º º º º º º º º º º º½º¾ ÓÖÑÙÐ Æ ÛØÓÒ º º º º º
( ) x y f(x, y) = ax
013 4 16 5 54 (03-5465-7040) nkiyono@mail.ecc.u-okyo.ac.jp hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
I, II 1, 2 ɛ-δ 100 A = A 4 : 6 = max{ A, } A A 10
1 2007.4.13. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 0. 1. 1. 2. 3. 2. ɛ-δ 1. ɛ-n
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2
II 1 3 2 5 3 7 4 8 5 11 6 13 7 16 8 18 2 1 1. x 2 + xy x y (1 lim (x,y (1,1 x 1 x 3 + y 3 (2 lim (x,y (, x 2 + y 2 x 2 (3 lim (x,y (, x 2 + y 2 xy (4 lim (x,y (, x 2 + y 2 x y (5 lim (x,y (, x + y x 3y
II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K
II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F
1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt
S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............
f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f
22 A 3,4 No.3 () (2) (3) (4), (5) (6) (7) (8) () n x = (x,, x n ), = (,, n ), x = ( (x i i ) 2 ) /2 f(x) R n f(x) = f() + i α i (x ) i + o( x ) α,, α n g(x) = o( x )) lim x g(x) x = y = f() + i α i(x )
DVIOUT-講
005-10-14 1 1 [1] [] [3] [4] (a + b) = a +ab + b [5] (a + b) 3 a 3 +a b + ab + a b +ab + b 3 a 3 +3a b +3ab + b 3 [6] (a + b) 4 (a + b) 5 [7] technology expand((a+b) n n =?) [8] technology n =6, 7, 8,
[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )
![[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )](/thumbs/85/91744595.jpg "[] x < T f(x), x < T f(x), < x < f(x) f(x) f(x) f(x + nt ) = f(x) x < T, n =, 1,, 1, (1.3) f(x) T x 2 f(x) T 2T x 3 f(x), f() = f(t ), f(x), f() f(t )")
1 1.1 [] f(x) f(x + T ) = f(x) (1.1), f(x), T f(x) x T 1 ) f(x) = sin x, T = 2 sin (x + 2) = sin x, sin x 2 [] n f(x + nt ) = f(x) (1.2) T [] 2 f(x) g(x) T, h 1 (x) = af(x)+ bg(x) 2 h 2 (x) = f(x)g(x)
1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :
9 ( ) 9 5 I II III A B (0 ) 5 I II III A B (0 ), 6 8 I II A B (0 ), 6, 7 I II A B (00 ) OAB A B OA = OA OB = OB A B : P OP AB Q OA = a OB = b () OP a b () OP OQ () a = 5 b = OP AB OAB PAB a f(x) = (log
68 A mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1
67 A Section A.1 0 1 0 1 Balmer 7 9 1 0.1 0.01 1 9 3 10:09 6 A.1: A.1 1 10 9 68 A 10 9 10 9 1 10 9 10 1 mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1 A.1. 69 5 1 10 15 3 40 0 0 ¾ ¾ É f Á ½ j 30 A.3: A.4: 1/10
(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (
![(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (](/thumbs/92/107866524.jpg "(, ) (, ) S = 2 = [, ] ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) ( ) (")
B 4 4 4 52 4/ 9/ 3/3 6 9.. y = x 2 x x = (, ) (, ) S = 2 = 2 4 4 [, ] 4 4 4 ( ) 2 ( ) 2 2 ( ) 3 2 ( ) 4 2 ( ) k 2,,, 4 4 4 4 4 k =, 2, 3, 4 S 4 S 4 = ( ) 2 + ( ) 2 2 + ( ) 3 2 + ( 4 4 4 4 4 4 4 4 4 ( (
2009 I 2 II III 14, 15, α β α β l 0 l l l l γ (1) γ = αβ (2) α β n n cos 2k n n π sin 2k n π k=1 k=1 3. a 0, a 1,..., a n α a
009 I II III 4, 5, 6 4 30. 0 α β α β l 0 l l l l γ ) γ αβ ) α β. n n cos k n n π sin k n π k k 3. a 0, a,..., a n α a 0 + a x + a x + + a n x n 0 ᾱ 4. [a, b] f y fx) y x 5. ) Arcsin 4) Arccos ) ) Arcsin
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
lecture
5 3 3. 9. 4. x, x. 4, f(x, ) :=x x + =4,x,.. 4 (, 3) (, 5) (3, 5), (4, 9) 95 9 (g) 4 6 8 (cm).9 3.8 6. 8. 9.9 Phsics 85 8 75 7 65 7 75 8 85 9 95 Mathematics = ax + b 6 3 (, 3) 3 ( a + b). f(a, b) ={3 (a
. sinh x sinh x) = e x e x = ex e x = sinh x 3) y = cosh x, y = sinh x y = e x, y = e x 6 sinhx) coshx) 4 y-axis x-axis : y = cosh x, y = s
. 00 3 9 [] sinh x = ex e x, cosh x = ex + e x ) sinh cosh 4 hyperbolic) hyperbola) = 3 cosh x cosh x) = e x + e x = cosh x ) . sinh x sinh x) = e x e x = ex e x = sinh x 3) y = cosh x, y = sinh x y =
r 1 m A r/m i) t ii) m i) t B(t; m) ( B(t; m) = A 1 + r ) mt m ii) B(t; m) ( B(t; m) = A 1 + r ) mt m { ( = A 1 + r ) m } rt r m n = m r m n B
1 1.1 1 r 1 m A r/m i) t ii) m i) t Bt; m) Bt; m) = A 1 + r ) mt m ii) Bt; m) Bt; m) = A 1 + r ) mt m { = A 1 + r ) m } rt r m n = m r m n Bt; m) Aert e lim 1 + 1 n 1.1) n!1 n) e a 1, a 2, a 3,... {a n
I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
Microsoft Word - 触ってみよう、Maximaに2.doc
i i e! ( x +1) 2 3 ( 2x + 3)! ( x + 1) 3 ( a + b) 5 2 2 2 2! 3! 5! 7 2 x! 3x! 1 = 0 ",! " >!!! # 2x + 4y = 30 "! x + y = 12 sin x lim x!0 x x n! # $ & 1 lim 1 + ('% " n 1 1 lim lim x!+0 x x"!0 x log x
1/1 lim f(x, y) (x,y) (a,b) ( ) ( ) lim limf(x, y) lim lim f(x, y) x a y b y b x a ( ) ( ) xy x lim lim lim lim x y x y x + y y x x + y x x lim x x 1
1/5 ( ) Taylor ( 7.1) (x, y) f(x, y) f(x, y) x + y, xy, e x y,... 1 R {(x, y) x, y R} f(x, y) x y,xy e y log x,... R {(x, y, z) (x, y),z f(x, y)} R 3 z 1 (x + y ) z ax + by + c x 1 z ax + by + c y x +
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h 0 f(a + h, b) f(a, b) h............................................................... ( ) f(x, y) (a, b) x A (a, b) x
i 6 3 ii 3 7 8 9 3 6 iii 5 8 5 3 7 8 v...................................................... 5.3....................... 7 3........................ 3.................3.......................... 8 3 35
名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト
名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim
OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P
4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e
40 6 y mx x, y 0, 0 x 0. x,y 0,0 y x + y x 0 mx x + mx m + m m 7 sin y x, x x sin y x x. x sin y x,y 0,0 x 0. 8 x r cos θ y r sin θ x, y 0, 0, r 0. x,
9.. x + y + 0. x,y, x,y, x r cos θ y r sin θ xy x y x,y 0,0 4. x, y 0, 0, r 0. xy x + y r 0 r cos θ sin θ r cos θ sin θ θ 4 y mx x, y 0, 0 x 0. x,y 0,0 x x + y x 0 x x + mx + m m x r cos θ 5 x, y 0, 0,
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,
[ ] 8 IC. y d y dx = ( dy dx ( p = dy p y dx ( ( ( 8 ( s8. 3 A A = ( A ( A (3 A P A P AP.3 π y(x = { ( 8 ( s8 x ( π < x x ( < x π y(x π π O π x ( 8 ( s83.4 f (x, y, z grad(f ( ( ( f f f grad(f = i + j
) 9 81
4 4.0 2000 ) 9 81 10 4.1 natural numbers 1, 2, 3, 4, 4.2, 3, 2, 1, 0, 1, 2, 3, integral numbers integers 1, 2, 3,, 3, 2, 1 1 4.3 4.3.1 ( ) m, n m 0 n m 82 rational numbers m 1 ( ) 3 = 3 1 4.3.2 3 5 = 2
1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2
filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin
6. Euler x
...............................................................................3......................................... 4.4................................... 5.5......................................
1.1 ft t 2 ft = t 2 ft+ t = t+ t 2 1.1 d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0
A c 2008 by Kuniaki Nakamitsu 1 1.1 t 2 sin t, cos t t ft t t vt t xt t + t xt + t xt + t xt t vt = xt + t xt t t t vt xt + t xt vt = lim t 0 t lim t 0 t 0 vt = dxt ft dft dft ft + t ft = lim t 0 t 1.1
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h f(a + h, b) f(a, b) h........................................................... ( ) f(x, y) (a, b) x A (a, b) x (a, b)
(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
数学Ⅲ立体アプローチ.pdf
Ⅲ Ⅲ DOLOR SET AMET . cos x cosx = cos x cosx = (cosx + )(cosx ) = cosx = cosx = 4. x cos x cosx =. x y = cosx y = cosx. x =,x = ( y = cosx y = cosx. x V y = cosx y = sinx 6 5 6 - ( cosx cosx ) d x = [
140 120 100 80 60 40 20 0 115 107 102 99 95 97 95 97 98 100 64 72 37 60 50 53 50 36 32 18 H18 H19 H20 H21 H22 H23 H24 H25 H26 H27 1 100 () 80 60 40 20 0 1 19 16 10 11 6 8 9 5 10 35 76 83 73 68 46 44 H11
, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x
![, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x](/thumbs/93/113428934.jpg ", 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x")
1 1.1 4n 2 x, x 1 2n f n (x) = 4n 2 ( 1 x), 1 x 1 n 2n n, 1 x n n 1 1 f n (x)dx = 1, n = 1, 2,.. 1 lim 1 lim 1 f n (x)dx = 1 lim f n(x) = ( lim f n (x))dx = f n (x)dx 1 ( lim f n (x))dx d dx ( lim f d
Chap10.dvi
=0. f = 2 +3 { 2 +3 0 2 f = 1 =0 { sin 0 3 f = 1 =0 2 sin 1 0 4 f = 0 =0 { 1 0 5 f = 0 =0 f 3 2 lim = lim 0 0 0 =0 =0. f 0 = 0. 2 =0. 3 4 f 1 lim 0 0 = lim 0 sin 2 cos 1 = lim 0 2 sin = lim =0 0 2 =0.
Chap9.dvi
.,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim
C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1
2006 7 18 1 2 C 2 2.1? 3x 2 + 2x + 5 = 0 (1) 1 2 7x + 4 = 0 (2) 1 1 x + x + 5 = 0 2 sin x x = 0 e x + x = 0 x = cos x (3) x + 5 + log x? 0.1% () 2.2 p12 3 x 3 3x 2 + 9x 8 = 0 (4) 1 [ ] 1/3 [ 2 1 ( x 1