41 · 2/3. Ҳосилаи функсияи мураккаб - Алгебра — синфи 10
Чаро қоида лозим аст
y = ( 8 x + 3 ) 2 y = (8x + 3)^2 y = ( 8 x + 3 ) 2 y = 64 x 2 + 48 x + 9 , y = 64x^2 + 48x + 9, y = 64 x 2 +
навбатӣ 41 · 3/3. Машқҳо бо ҳосилаи функсияи мураккаб 1 соат
48
x
+
9 ,
y ′ = 128 x + 48. y' = 128x + 48. y ′ = 128 x + 48. ( 2 x + 4 ) 100 (2x + 4)^{100} ( 2 x + 4 ) 100 y = u 2 , y = u^2, y = u 2 , u = 8 x + 3 u = 8x + 3 u = 8 x + 3 y ′ ( u ) = 2 u , y'(u) = 2u, y ′ ( u ) = 2 u , u ′ ( x ) = 8. u'(x) = 8. u ′ ( x ) = 8. y y y u u u 2 u 2u 2 u u u u x x x 8 8 8 y y y x x x 2 u ⋅ 8 2u \cdot 8 2 u ⋅ 8 y ′ ( x ) = 2 u ⋅ 8 = 16 ( 8 x + 3 ) = 128 x + 48. y'(x) = 2u \cdot 8 = 16(8x + 3) = 128x + 48. y ′ ( x ) = 2 u ⋅ 8 = 16 ( 8 x + 3 ) = 128 x + 48.
Теорема Исбот. Ба x 0 x_0 x 0 Δ x ≠ 0 \Delta x \ne 0 Δ x = 0 Δ u = f ( x 0 + Δ x ) − f ( x 0 ) \Delta u = f(x_0 + \Delta x) - f(x_0) Δ u = f ( x 0 + Δ x ) − f ( x 0 ) Δ F = g ( u 0 + Δ u ) − g ( u 0 ) . \Delta F = g(u_0 + \Delta u) - g(u_0). Δ F = g ( u 0 + Δ u ) − g ( u 0 ) .
Функсияи Q ( t ) Q(t) Q ( t ) t ≠ 0 t \ne 0 t = 0 Q ( t ) = g ( u 0 + t ) − g ( u 0 ) t , Q(t) = \dfrac{g(u_0 + t) - g(u_0)}{t}, Q ( t ) = t g ( u 0 + t ) − g ( u 0 ) , t = 0 t = 0 t = 0 Q ( 0 ) = g ′ ( u 0 ) . Q(0) = g'(u_0). Q ( 0 ) = g ′ ( u 0 ) . Q ( t ) → g ′ ( u 0 ) Q(t) \to g'(u_0) Q ( t ) → g ′ ( u 0 ) t → 0. t \to 0. t → 0. Δ u = 0 \Delta u = 0 Δ u = 0 Δ F = Q ( Δ u ) ⋅ Δ u . \Delta F = Q(\Delta u) \cdot \Delta u. Δ F = Q ( Δ u ) ⋅ Δ u . Δ F Δ x = Q ( Δ u ) ⋅ Δ u Δ x . \frac{\Delta F}{\Delta x} = Q(\Delta u) \cdot \frac{\Delta u}{\Delta x}. Δ x Δ F = Q ( Δ u ) ⋅ Δ x Δ x → 0 \Delta x \to 0 Δ x → 0 f f f Δ u → 0 , \Delta u \to 0, Δ u → 0 , Q ( Δ u ) → g ′ ( u 0 ) ; Q(\Delta u) \to g'(u_0); Q ( Δ u ) → g ′ ( u 0 ) ; Δ u Δ x → f ′ ( x 0 ) . \dfrac{\Delta u}{\Delta x} \to f'(x_0). Δ x Δ u → f ′ ( x 0 ) . F ′ ( x 0 ) = g ′ ( u 0 ) ⋅ f ′ ( x 0 ) . ■ F'(x_0) = g'(u_0) \cdot f'(x_0). \qquad \blacksquare F ′ ( x 0 ) = g ′ ( u
Мисолҳо Мисоли 1. Ҳосилаи функсияҳоро меёбем.
а) y = ( 9 x 2 − 1 ) 4 . y = (9x^2 - 1)^4. y = ( 9 x 2 − 1 ) 4 . g ( u ) = u 4 , g(u) = u^4, g ( u ) = u 4 , u = 9 x 2 − 1 u = 9x^2 - 1 u = 9 x 2 − 1 g ′ ( u ) = 4 u 3 , g'(u) = 4u^3, g ′ ( u ) = 4 u 3 , u ′ = 18 x . u' = 18x. u ′ = 18 x . y ′ = 4 ( 9 x 2 − 1 ) 3 ⋅ 18 x = 72 x ( 9 x 2 − 1 ) 3 . y' = 4(9x^2 - 1)^3 \cdot 18x = 72x\,(9x^2 - 1)^3. y ′ = 4 ( 9 x 2 − 1 ) 3 ⋅ 18 x =
б) y = 5 x 3 − 3 x 2 + x − 9 . y = \sqrt{5x^3 - 3x^2 + x - 9}. y = 5 x 3 − 3 x 2 + x − 9 . g ( u ) = u , g(u) = \sqrt{u}, g ( u ) = u , g ′ ( u ) = 1 2 u , g'(u) = \dfrac{1}{2\sqrt{u}}, g ′ ( u ) = 2 u u ′ = 15 x 2 − 6 x + 1 : u' = 15x^2 - 6x + 1: u ′ = 15 x 2 − 6 x + 1 : y ′ = 15 x 2 − 6 x + 1 2 5 x 3 − 3 x 2 + x − 9 . y' = \frac{15x^2 - 6x + 1}{2\sqrt{5x^3 - 3x^2 + x - 9}}. y ′ = 2 5 x 3 − 3 x
Ду ҳолати махсус зуд-зуд вомехӯрад.
Мисоли 2. y = ( 2 x + 1 ) 3 ( x 2 + x + 1 ) 2 . y = \dfrac{(2x + 1)^3}{(x^2 + x + 1)^2}. y = ( x 2 + x + 1 ) 2 ( 2 x + 1 ) 3 . y ′ = 3 ⋅ 2 ( 2 x + 1 ) 2 ( x 2 + x + 1 ) 2 − ( 2 x + 1 ) 3 ⋅ 2 ( x 2 + x + 1 ) ( 2 x + 1 ) ( x 2 + x + 1 ) 4 . y' = \frac{3 \cdot 2\,(2x + 1)^2 (x^2 + x + 1)^2 - (2x + 1)^3 \cdot 2(x^2 + x + 1)(2x + 1)}{(x^2 + x + 1)^4}. y ′ = ( x ( 2 x + 1 ) 2 ( x 2 + x + 1 ) (2x + 1)^2 (x^2 + x + 1) ( 2 x + 1 ) 2 ( x 2 + x + 1 ) y ′ = ( 2 x + 1 ) 2 [ 6 ( x 2 + x + 1 ) − 2 ( 2 x + 1 ) 2 ] ( x 2 + x + 1 ) 3 = − 2 ( x 2 + x − 2 ) ( 2 x + 1 ) 2 ( x 2 + x + 1 ) 3 . y' = \frac{(2x + 1)^2 \bigl[\,6(x^2 + x + 1) - 2(2x + 1)^2\bigr]}{(x^2 + x + 1)^3} = -\frac{2\,(x^2 + x - 2)(2x + 1)^2}{(x^2 + x + 1)^3}. y ′ =
Мисоли 3. f ( x ) = 3 + 5 x 3 . f(x) = \sqrt{3 + 5x^3}. f ( x ) = 3 + 5 x 3 . g ( u ) = u , g(u) = \sqrt{u}, g ( u ) = u , u = 3 + 5 x 3 , u = 3 + 5x^3, u = 3 + 5 x 3 , u ′ = 15 x 2 : u' = 15x^2: u ′ = 15 x 2 : f ′ ( x ) = 1 2 u ⋅ 15 x 2 = 15 x 2 2 3 + 5 x 3 . f'(x) = \frac{1}{2\sqrt{u}} \cdot 15x^2 = \frac{15x^2}{2\sqrt{3 + 5x^3}}. f ′ ( x ) = 2 u
Кор дар синф
Машқҳо
Дар синф 1. Ҳосилаи функсияро ёбед:
а) ( 11 + x ) 21 ; (11 + x)^{21}; ( 11 + x ) 21 ;
б) ( 9 x + 23 ) − 4 ; (9x + 23)^{-4}; ( 9 x + 23 ) − 4 ;
в) ( 0,1 x − 1 ) − 10 ; (0{,}1x - 1)^{-10}; ( 0 , 1 x − 1 ) − 10 ;
г) x + 3,2 ; \sqrt{x + 3{,}2}; x + 3 , 2 ;
д) 9 − 2 x ; \sqrt{9 - 2x}; 9 − 2 x ;
е) 3 x − 91 ; \sqrt{3x - 91}; 3 x − 91 ;
ж) x 2 + 13 ; \sqrt{\dfrac{x}{2} + 13}; 2 x + 13 ;
з) ( a x + b ) n ; (ax + b)^n; ( a x + b ) n ;
и) ( a x + b ) − n . (ax + b)^{-n}. ( a x + b ) − n .
2. Ҳосилаи функсияро ёбед:
а) 5 x 2 − 27 ; \sqrt{5x^2 - 27}; 5 x 2 − 27 ;
б) x 2 + 10 x − 61 ; \sqrt{x^2 + 10x - 61}; x 2 + 10 x − 61 ;
в) x − 1 + x 2 + 1 ; \sqrt{x - 1} + \sqrt{x^2 + 1}; x − 1 + x 2 + 1
г) 8 x 3 + 5 ; \sqrt{8x^3 + 5}; 8 x 3 + 5 ;
д) 0,25 x 4 + 2 + x 3 ; \sqrt{0{,}25x^4 + 2} + \sqrt{x^3}; 0 , 25 x 4 + 2 + x
е) a x k + b x + c . \sqrt{ax^k + bx + c}. a x k + b x + c .
Дар хона 3. Ҳосилаи функсияро ёбед:
а) ( 3 x − 10 ) 23 ; (3x - 10)^{23}; ( 3 x − 10 ) 23 ;
б) ( x − 2 ) 70 ; (x - 2)^{70}; ( x − 2 ) 70 ;
в) ( 7 x − 4 ) 101 . (7x - 4)^{101}. ( 7 x − 4 ) 101 .
4. Ҳосилаи функсияро ёбед:
а) 1 ( 3 + 5 x ) 11 ; \dfrac{1}{(3 + 5x)^{11}}; ( 3 + 5 x ) 11 1 ;
б) 4 ( 1 + 2 x ) 29 ; \dfrac{4}{(1 + 2x)^{29}}; ( 1 + 2 x ) 29 4 ;
в) − 1 ( 3 x + 2 ) 99 . -\dfrac{1}{(3x + 2)^{99}}. − ( 3 x + 2 ) 99 1 .
5. Ҳосилаи функсияро ёбед:
а) 25 − x 3 ; \sqrt{25 - x^3}; 25 − x 3 ;
б) 1 9 − x 2 ; \dfrac{1}{\sqrt{9 - x^2}}; 9 − x 2 1 ;
в) 2 + 1 x . \sqrt{2 + \dfrac{1}{x}}. 2 + x 1 .
6. Ҳосилаи функсияро ёбед:
а) x 2 + 3 x ; x^{\sqrt{2}} + 3x; x 2 + 3 x ;
б) ( x + 1 ) 3 ; (x + 1)^{\sqrt{3}}; ( x + 1 ) 3 ;
в) x 5 − 8. x^{\sqrt{5}} - 8. x 5 − 8.
Вазифаи хонагӣ Дар ин дарс
Ман медонам формулаи ҳосилаи функсияи мураккаб ва исботи он.Ман метавонам ҳосилаи ( k x + b ) n \left(kx + b\right)^n ( k x + b ) n a x 2 + b x + c \sqrt{ax^2 + bx + c} a x 2 + b x + c Ман мефаҳмам , ки чаро суръатҳои тағйирёбӣ зарб мешаванд. Такрор (2 дақиқа)
( x 5 ) ′ , \left(x^5\right)', ( x 5 ) ′ , ( x ) ′ , \left(\sqrt{x}\right)', ( x ) ′ , ( 1 x ) ′ \left(\dfrac{1}{x}\right)' ( x 1 ) ′ y = ( 2 x + 1 ) 2 y = (2x + 1)^2 y = ( 2 x + 1 ) 2 y = 4 − x 2 y = 4 - x^2 y = 4 − x 2 u = 4 − x 2 u = 4 - x^2 u = 4 − x 2 u \sqrt{u} Ҷавобҳо
5 x 4 , 5x^4, 5 x 4 , 1 2 x , \dfrac{1}{2\sqrt{x}}, 2 x Вазъият барои андешаронӣ y = ( 2 x + 4 ) 100 y = (2x + 4)^{100} y = ( 2 x + 4 ) 100
Теорема Агар f f f x 0 x_0 x 0 g g g u 0 = f ( x 0 ) u_0 = f(x_0)
Натиҷаи 1 Агар y = f ( k x + b ) y = f(kx + b) y = f ( k x + b ) k , k, k , b b b y ′ = k ⋅ f ′ ( k x + b ) . y' = k \cdot f'(kx + b).
Натиҷаи 2 Агар y = f ( a x 2 + b x + c ) y = f(ax^2 + bx + c) y = f ( a x 2 + b x + c ) y ′ = ( 2 a x + b ) ⋅ f ′ ( a x 2 + b x + c ) . y' = (2ax + b) \cdot f'(ax^2 + bx + c).
Кори дунафара Ҳосилаи y = ( 8 x + 3 ) 2 y = (8x + 3)^2 y = ( 8 x + 3 ) 2 y = ( 4 x + 7 ) 3 y = (4x + 7)^3 y =
Ман дар ин дарс чӣ омӯхтам? Бо ҷумлаи худ шарҳ диҳед: чаро ҳангоми ҳосилагирии ( 8 x + 3 ) 2 (8x + 3)^2 ( 8 x + 3 ) 2 2 u 2u 2 u 8 8 8
( ( x 2 + 1 ) 3 ) ′ = 3 ( x 2 + 1 ) 2 . \left((x^2 + 1)^3\right)' = 3(x^2 + 1)^2. ( ( x 2 + 1 ) 3 ) ′ = 3 ( x
( x 2 + 1 ) ′ \left(\sqrt{x^2 + 1}\right)' ( x 2 + 1 ) ′
Як ҷавобро интихоб кунед A
B C D Қадамҳои ёфтани ҳосилаи y = ( 4 x − 7 ) 3 y = (4x - 7)^3 y = ( 4 x − 7 ) 3
Унсурҳоро бо тартиби дуруст ҷойгир кунед 1 Зарб мекунем ва u u u y ′ = 3 ( 4 x − 7 ) 2 ⋅ 4 y' = 3(4x - 7)^2 \cdot 4 y
2
3
4
Δ u
.
0
)
⋅
f ′ ( x 0 ) . ■
72 x ( 9 x 2 −
1 ) 3 .
1
,
2
+
x
−
9
15 x 2 − 6 x + 1
.
2
+
x
+
1
) 4
3 ⋅ 2 ( 2 x + 1 ) 2 ( x 2 + x + 1 ) 2 − ( 2 x + 1 ) 3 ⋅ 2 ( x 2 + x + 1 ) ( 2 x + 1 )
.
( x 2 + x + 1 ) 3
( 2 x + 1 ) 2 [ 6 ( x 2 + x + 1 ) − 2 ( 2 x + 1 ) 2 ]
=
− ( x 2 + x + 1 ) 3 2 ( x 2 + x − 2 ) ( 2 x + 1 ) 2 .
1
⋅
15 x 2 =
;
3
;
1
,
− 1 x 2 . -\dfrac{1}{x^2}. − x 2 1 . 4 x 2 + 4 x + 1 , 4x^2 + 4x + 1, 4 x 2 + 4 x + 1 , u
0
=
f ( x 0 )
F ( x ) = g [ f ( x ) ] F(x) = g[f(x)] F ( x ) = g [ f ( x )] F ′ ( x 0 ) = g ′ [ f ( x 0 ) ] ⋅ f ′ ( x 0 ) . F'(x_0) = g'\bigl[f(x_0)\bigr] \cdot f'(x_0). F ′ ( x 0 ) = g ′ [ f ( x 0 ) ] ⋅ f ′ ( x 0 ) . y ′ ( x ) = y ′ ( u ) ⋅ u ′ ( x ) . y'(x) = y'(u) \cdot u'(x). y ′ ( x ) = y ′ ( u ) ⋅ u ′ ( x ) .
y ′
=
k ⋅
f ′ ( k x +
b ) .
y ′ =
( 2 a x +
b ) ⋅
f ′ ( a x 2 +
b x +
c ) .
( 4 x +
7 ) 3
Ҷавоб 128 x + 48 128x + 48 128 x + 48 y ′ = 3 ( 4 x + 7 ) 2 ⋅ 4 = 12 ( 4 x + 7 ) 2 . y' = 3(4x + 7)^2 \cdot 4 = 12(4x + 7)^2. y ′ = 3 ( 4 x + 7 ) 2 ⋅ 4 = 12 ( 4 x + 7 ) 2 .
2
+
1 ) 2 .
x x 2 + 1 \dfrac{x}{\sqrt{x^2 + 1}} 2 x x 2 + 1 2x\sqrt{x^2 + 1} 2 x x 2 + 1
1 2 x 2 + 1 \dfrac{1}{2\sqrt{x^2 + 1}} 2 x 2 + 1 1
2 x x 2 + 1 \dfrac{2x}{\sqrt{x^2 + 1}} x 2 + 1 2 x
′
=
3 ( 4 x −
7 ) 2 ⋅
4
Ҷавобро менависем: y ′ = 12 ( 4 x − 7 ) 2 y' = 12(4x - 7)^2 y ′ = 12 ( 4 x − 7 ) 2
Ҳосилаҳоро меёбем: y ′ ( u ) = 3 u 2 , y'(u) = 3u^2, y ′ ( u ) = 3 u 2 , u ′ ( x ) = 4 u'(x) = 4 u ′ ( x ) = 4
Менависем: y = u 3 , y = u^3, y = u 3 , u = 4 x − 7 u = 4x - 7 u = 4 x − 7