高等学校学習指導要領
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- あきたけ つちかね
- 9 years ago
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No. 1 2 3 1 4 310 1 5 311 7 1 6 311 1 7 2 8 2 9 1 10 2 11 2 12 2 13 3 14 3 15 3 16 3 17 2 18 2 19 3 1 No. 20 4 21 4 22 4 23 4 25 4 26 4 27 4 28 4 29 2760 4 30 32 6364 4 36 4 37 4 39 4 42 4 43 4 44 4 46
lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d
lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3
, 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,
2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)
1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a
(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {
7 4.., ], ], ydy, ], 3], y + y dy 3, ], ], + y + ydy 4, ], ], y ydy ydy y y ] 3 3 ] 3 y + y dy y + 3 y3 5 + 9 3 ] 3 + y + ydy 5 6 3 + 9 ] 3 73 6 y + y + y ] 3 + 3 + 3 3 + 3 + 3 ] 4 y y dy y ] 3 y3 83 3
() 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 ()
1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2
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
( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +
(.. C. ( d 5 5 + C ( d d + C + C d ( d + C ( ( + d ( + + + d + + + + C (5 9 + d + d tan + C cos (sin (6 sin d d log sin + C sin + (7 + + d ( + + + + d log( + + + C ( (8 d 7 6 d + 6 + C ( (9 ( d 6 + 8 d
Gmech08.dvi
51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r
f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f
208 3 28. f fd f Df 関数 接線 D f f 0 f f f 2 f f f f f 3 f lim f f df 0 d 4 f df d 3 f d f df d 5 d c 208 2 f f t t f df d 6 d t dt 7 f df df d d df dt lim f 0 t df d d dt d t 8 dt 9.2 f,, f 0 f 0 lim 0 lim
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
( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (
6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b
grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = 0 g (0) g (0) (31) grad φ(p ) p grad φ φ (P, φ(p )) xy (x, y) = (ξ(t), η(t)) ( )
2 9 2 5 2.2.3 grad φ(p ) φ P grad φ(p ) p P p φ P p l t φ l t = g () g () (3) grad φ(p ) p grad φ φ (P, φ(p )) y (, y) = (ξ(t), η(t)) ( ) ξ (t) (t) := η (t) grad f(ξ(t), η(t)) (t) g(t) := f(ξ(t), η(t))
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1 1 1. 2. 3. 2 2 1 (5/6) 4 =0.517... 5/6 (5/6) 4 1 (5/6) 4 1 (35/36) 24 =0.491... 0.5 2.7 3 1 n =rand() 0 1 = rand() () rand 6 0,1,2,3,4,5 1 1 6 6 *6 int() integer 1 6 = int(rand()*6)+1 1 4 3 500 260 52%
高等学校学習指導要領解説 数学編
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Gmech08.dvi
63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)
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IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r
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38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
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
m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d
m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )
n n (n) (n) (n) (n) n n ( n) n n n n n en1, en ( n) nen1 + nen nen1, nen ( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( ) ( n) Τ n n n ( n) n + n ( n) (n) n + n n n n n n n n
1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1
ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD
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
さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1
... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =
< 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) =
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
(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
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DT-870/5100 &DT-5042RFB ...3 1-1...3 1-1...6 1-3...16 2....17...21 3-1...21 3-2...21 3-2...22 3-3...23 3-4...24...25 4-1....25 4-2...27 4-3...28 4-4...33 4-5...36...37 5-1....39 5-2...40 5-3...43...49
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
( ) x y f(x, y) = ax
013 4 16 5 54 (03-5465-7040) [email protected] 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
(ii) (iii) z a = z a =2 z a =6 sin z z a dz. cosh z z a dz. e z dz. (, a b > 6.) (z a)(z b) 52.. (a) dz, ( a = /6.), (b) z =6 az (c) z a =2 53. f n (z
B 4 24 7 9 ( ) :,..,,.,. 4 4. f(z): D C: D a C, 2πi C f(z) dz = f(a). z a a C, ( ). (ii), a D, a U a,r D f. f(z) = A n (z a) n, z U a,r, n= A n := 2πi C f(ζ) dζ, n =,,..., (ζ a) n+, C a D. (iii) U a,r
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
...............y.\....07..
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知能科学:ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4
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高校生の就職への数学II
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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
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dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp
+ P (x)y = Q(x) (1) = P (x)y + Q(x) P (x), Q(x) y Q(x) 0 homogeneous = P (x)y 1 y = P (x) log y = P (x) + C y = C exp{ P (x) } = C e R P (x) 5.1 + P (x)y = 0 (2) y = C exp{ P (x) } = Ce R P (x) (3) αy
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But nothing s unconditional, The Bravery R R >0 = (0, ) ( ) R >0 = (0, ) f, g R >0 f (0, R), R >
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Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0
Tips KENZOU 28 7 6 P no problem 2 w = f(z) z w w z w = (z z ) b b w = log (z z ) z = z 2π 2 z = z w = z /2 z = re iθ θ (z = ) 2π 4π 2 θ θ 2π 4π z r re iθ re i2π = r re i4π = r w r re iθ/2 re iπ = r re
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP
2010 IA ε-n I 1, 2, 3, 4, 5, 6, 7, 8, ε-n 1 ε-n ε-n? {a n } n=1 1 {a n } n=1 a a {a n } n=1 ε ε N N n a n a < ε
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f : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し 原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y
017 8 10 f : R R f(x) = x n + x n 1 + 1, f(x) = sin 1, log x x n m :f : R n R m z = f(x, y) R R R R, R R R n R m R n R m R n R m f : R R f (x) = lim h 0 f(x + h) f(x) h f : R n R m m n M Jacobi( ) m n
7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a
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1 1 1 1 1 1 2 f z 2 C 1, C 2 f 2 C 1, C 2 f(c 2 ) C 2 f(c 1 ) z C 1 f f(z) xy uv ( u v ) = ( a b c d ) ( x y ) + ( p q ) (p + b, q + d) 1 (p + a, q + c) 1 (p, q) 1 1 (b, d) (a, c) 2 3 2 3 a = d, c = b
1 I p2/30
I I p1/30 1 I p2/30 1 ( ) I p3/30 1 ( ), y = y() d = f() g(y) ( g(y) = f()d) (1) I p4/30 1 ( ), y = y() d = f() g(y) ( g(y) = f()d) (1) g(y) = f()d I p4/30 1 ( ), y = y() d = f() g(y) ( g(y) = f()d) (1)
φ s i = m j=1 f x j ξ j s i (1)? φ i = φ s i f j = f x j x ji = ξ j s i (1) φ 1 φ 2. φ n = m j=1 f jx j1 m j=1 f jx j2. m
2009 10 6 23 7.5 7.5.1 7.2.5 φ s i m j1 x j ξ j s i (1)? φ i φ s i f j x j x ji ξ j s i (1) φ 1 φ 2. φ n m j1 f jx j1 m j1 f jx j2. m j1 f jx jn x 11 x 21 x m1 x 12 x 22 x m2...... m j1 x j1f j m j1 x
B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:
B. 41 II: ;; 4 B [] S 1 S S 1 S.1 O S 1 S 1.13 P 3 P 5 7 P.1:.13: 4 4.14 C d A B x l l d C B 1 l.14: AB A 1 B 0 AB 0 O OP = x P l AP BP AB AP BP 1 (.4)(.5) x l x sin = p l + x x l (.4)(.5) m d A x P O
[ ] 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)
重力方向に基づくコントローラの向き決定方法
( ) 2/Sep 09 1 ( ) ( ) 3 2 X w, Y w, Z w +X w = +Y w = +Z w = 1 X c, Y c, Z c X c, Y c, Z c X w, Y w, Z w Y c Z c X c 1: X c, Y c, Z c Kentaro [email protected] 1 M M v 0, v 1, v 2 v 0 v
3 1 1 BCA ACD HP A AB BC ABC ONP x AM, CN x 30 DM DM! CN CN AM AMD 10 1 AB AC
3 BCA ACD HP A AB BC ABC ONP x AM, CN x 30 DM DM!CN CN AM AMD 0 AB AC AMD3AMB3030 x x x 0,60,50,30,30 ABCD AC, BD ABD, DBC, BCA, ACD BDA x BDC 80 AB AD ABD sin x sin AD AC ACD sin sin x AC ABC sin AB sin
2 x x, y, z x,, z F c : x x x cos y sin z z 8 F F F F F x x x F x x F 9 F c J Fc J Fc x x x y y y cos sin 0 sin cos 0 0 0, J Fc 0 J Fc t x /x J Fc,, z
The Austalian National Univesity 9 5 F n n 0,, 2, F F ě i F f F g F g 2 ij F F x. ě i F F x i i, 2, 3 2 f i ě i f f x i. 3 g ij ě i g ě j g x i ě j g j x i. 4 g ij g x g 2 x g 3 x g x 2 g 2 x 2 g 3 x 2
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
