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Review of Chapter 3 - 已學過的 rules( 回顧 )-. 朝陽科技大學 資訊管理系 李麗華 教授. 3.2 (p.115). Derivative of a constant in zero (p.115 上 ) Power Rule (p.115 下 ). n : real. (p.116 中 ). 3.2 (p.115). Sam & Difference Rules (p.117 下 ). 即. 即. 3.2 (p.115). Product Rule (p.140 上 ) Quotient Rule (p.142 中 ) - PowerPoint PPT Presentation
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Review of Chapter 3 - 已學過的 rules( 回顧 )-
朝陽科技大學資訊管理系李麗華 教授
2
3.2 (p.115)
1. Derivative of a constant in zero (p.115 上 )
2. Power Rule (p.115 下 )
3.
0dc
dx
1( )n ndx nx
dx n: real
[ ( )] ( )d dc f x c f x
dx dx (p.116 中 )
3
3.2 (p.115)
4. Sam & Difference Rules (p.117 下 )
即
即
[ ( ) ( )] ( ) ( )d d df x g x f x g x
dx dx dx
'( ) '( )f x g x
[ ( ) ( )] ( ) ( )d d df x g x f x g x
dx dx dx
'( ) '( )f x g x
4
3.2 (p.115)
5. Product Rule (p.140 上 )
6. Quotient Rule (p.142 中 )
7. Chain Rule (p.140 下 )
( ) ' 'du v u v v u
dx
2
' '( )
d u u v uv
dx v v
dy dy du
dx du dx
5
3.2 (p.115)
8. General Power Rule
1n nd duu n u
dx dx
or 1'( ) [ ( )] '( )nf x n u x u x
6
3.5 The Product ( 積 ) & Quotient Rule ( 商 )
1. 由於函數與函數間的 、 、 、 和冪次等諸多變化,茲將為分的法則分別介紹。
2. 已在前面學了和、差法則,即
然而積與商法則都不是可以分開帶入計算的。
/
[ ( ) ( )]df x g x
dx
( ) ( )d df x g x
dx dx
EX:2( )f x x ( )g x x , 則 [ ( ) ( )]' '( ) '( )f x g x f x g x
3 2( ) ' 3x x 2 1 2x x
7
3.5 The Product ( 積 ) & Quotient Rule ( 商 )
3. Product Rule
Let ( )u u x , ( )v v x then ( ) ' 'du v u v u v
dx
即 ( ) [ ( )] [ ( )] ( )d d
f x g x f x g xdx dx
,
( 或 ( ) ' 'df g f g f g
dx )
8
3.5 The Product ( 積 ) & Quotient Rule ( 商 )
3. Product Rule
proof:已知0
( ) ( ) ( ) ( )[ ( ) ( )] lim
x
d u x x v x x u x v xu x v x
dx x
加入一個[ ( ) ( ) ( ) ( )]u x v x x u X v x x
加入項0
( ( ) ( ) ( ) ( )) ( ( ) ( ) ( ) ( ))limx
u x x v x x u x v x x u x v x x u x v x
x
拆兩項0 0
( ( ) ( ) ( ) ( )) ( ( ) ( ) ( ) ( ))lim limx x
u x x v x x u x v x x u x v x x u x v x
x x
提出共同項0 0
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )lim limx x
u x x v x x u x v x x u x v x x u x v x
x x
拆0 0 0 0
( ) ( ) ( ) ( )[ lim ] lim ( ) lim ( ) [ lim ]x x x x
u x x u x v x x v xv x x u x
x x
'( ) ( ) ( ) '( )u x v x u x v x 得証
9
3.5 The Product & Quotient Rule 範例
EX : 2( ) (3 5)( 7 )f x x x x '( )f x, 求
sol :2 2'( ) (3 5) ( 7 ) ' (3 5) '( 7 )f x x x x x x x
2(3 5) (2 7) (3)( 7 )x x x x 2 26 21 35 10 3 21x x x x x
9 2 32 35x x
EX :
sol :
1( ) (2 1)(1 )f x x
x '( )f x, 求
1 1'( ) (2 1) '(1 ) (2 1)(1 ) 'f x x x
x x
212(1 ) (2 1) ( )x x
x
2 2
2 2 2
2 2 1 2 2 2 1 2 12
x x x x x
x x x x
10
上台練習
EX1 :
EX2 :
3 2( ) ( 1)( 3)f x x x
3 2( ) (1 )(1 )f x x x
EX3 :2( 5)(1 2 )y x x
EX4 :1
(4 1)(1 )y xx
11
3.5 The Product ( 積 ) & Quotient Rule ( 商 )
4. Quotient Rule
2
' '( )
d u u v uv
dx v v
EX :
sol :
5 1( )
1 2
xf x
x
, find the derivative of ( )f x
2
(5 1) '(1 2 ) (5 1)(1 2 ) ''( )
(1 2 )
x x x xf x
x
2 2
5 10 10 2 7
(1 2 ) (1 2 )
x x
x x
EX :
sol :
1( )
4 3
xf x
x
2 2 2
( 1) '(4 3) ( 1)(4 3) 4 3 4 4 7'( )
(4 3) (4 3) (4 3)
x x x x x xf x
x x x
, find the derivative of ( )f x
12
上台練習
EX1 :
EX2 :
EX3 :
EX4 :
24( )
1 5
t ts t
t
4 2
( )3
tf t
t
12
4 1( )
xf x
x
2
3 1( )
2
xf x
x x
13
3.6 The Chain Rule
• 前面已學 power rule ,即 ,但這個法則並不能直接套在 這樣的式子,即
,若將 視為另一個函數,即 ,故 ,那麼微分應該是 ,即 chain rule 。
• Chain Rule :若 y is func. of u and u is func. of x
1n ndx n x
dx
2 3( 1)x 2 3 2 2( 1) 3( 1)
dx x
dx 2 3( 1)x
2 1u x 3( )f x udy dy du
dx du dx
dy dy du
dx du dx
14
3.6 The Chain Rule 範例
EX :
sol :
若 2 8( 1)y x
let 2 1u x 8y u
7 2 7 28 ( 1) 8 2 16 ( 1)dy dy du
u x u x x xdx du dx
EX :
sol :
2 4y x 求dy
dx
let 2 4u x 12y u
1 12 22
2
1(2 ) ( 4)
2 4
dy dy du xu x x x
dx du dx x
15
3.6 The Chain Rule
• 因此若前面的 power rule 中的 x是另一個函數的話,則可修改如下:– General Power Rule
1n nd duu nu
dx dx 1 'ndy du
nu udu dx
u is a differentiable function of x and n is a real number
16
上台練習
EX1 :
EX2 :
EX3 :
EX4 :
2 3
1( )
( 5)f x
x
4 3( ) (1 )f x x
5( ) (3 )f x x
1( )
6 5f x
x
, {
2 5u x
let
∴
13( )f x u
, {41u x
3( )f x u
, {
, {
3u x
6 5u x
5( )f x u
12( )f x u
, 求
, 求
, 求
, 求
'( )f x
'( )f x
'( )f x
'( )f x
17
综合練習
EX1 :
sol :
2 2 2 5( ) (4 1) ( 3)f x x x
let 2 2(4 1)u x , 2 5( 3)v x 2 2 2 5 2 2 2 5'( ) (4 1) [( 3) ]' [(4 1) ]' ( 3)f x x x x x 2 2 2 4 2 2 5(4 1) [10 ( 3) ] [2(4 1)8 ] ( 3)x x x x x x
2 2 2 4 2 2 510 (4 1) ( 3) 16 (4 1) ( 3)x x x x x x
18
综合練習
EX2 :
sol :
4 3( ) (2 3) ( 7)f t t t , 求 '( )f t
4 3 4 3'( ) (2 3) [( 7) ]' [(2 3) ]'( 7)f t t t t t 4 2 3 3(2 3) 3( 7) 1 [4(2 3) 2] ( 7)t t t t
4 2 3 33(2 3) ( 7) 8(2 3) ( 7)t t t t
EX3 :
sol :
同理利用 Quotient Rule 應用 ,
5(2 1)( )
3 1
xf x
x
, 求 '( )f x
5 5
2
[(2 1) ]'(3 1) (2 1) (3)'( )
(3 1)
x x xf x
x
4 5
2
[5(2 1) 2](3 1) 3(2 1)
(3 1)
x x x
x
5
2
10(2 1)(3 1) 3(2 1)
(3 1)
x x x
x
19
上台練習
EX1 :
EX2 :
EX3 :
3( 4)
1
xy
x
3( 3)(2 1)y x x
32(2 5)(2 1)y x x
20
3.7 High-Order Derivatives ( 高階導函數 )
• 前面所學均為一次微分,即 ,而高階即指多階微分之意,例:
'( )f x
''( )f x '''( )f x, …
寫法: 'y 'f dy
dxor xD y
''y ''f 2
2
d y
dxor 2
xD y
'''y '''f 3
3
d y
dxor 3
xD y
( * ) 計算式即逐次對前一個微分結果再做微分即可得高一階的微分
21
3.7 High-Order Derivatives ( 高階導函數 )
EX : 5y x
則dy
dxor 'y or 4'( ) 10f x x
3''( ) 40f x x2'''( ) 120f x x
(4) ( ) 240f x x
EX : 3 2( ) 7f x x x x , 求 (3)f , (4)f
2'( ) 2 7 1f x x x
''( ) 4 7f x x (3) ( ) 4f x ,
(4) ( ) 0f x
22
3.7 High-Order Derivatives ( 高階導函數 )
• 二階微分即 ,我們通常稱為一階函數的變化率,日常生活中常見的例子即”加速度”(Acceleration) 。
''( )f x
23
3.7 High-Order Derivatives ( 高階導函數 )
EX : 若一球往上丟之距離公式為 ,則請求出這個球在216 80S t t
3t 的速度及加速度。
sol : ' 32 80S t
'' 32S
∴ 時的速度為3t '(3) 32(3) 80 96 80 16S
3t 的加速度為 32
24
3.7 High-Order Derivatives ( 高階導函數 )
EX : 若一公司生產物品的成本為 ,請求出當2( ) 800 50 0.04C x x x
35x 的邊際成本 (marginal cost) 的 rate of change 。
sol : marginal cost 即求 ,而求 marginal cost 的 rate of change'( ) 0.08 50C x x
即 ( 即遞減的固定變化量 ) 。''( ) 0.08C x
25
3.8 Implicit Differentiation ( 隱微分 )
1. 若遇 這類式子,因為無法寫出所以無法直接套用所學的微分方法。對這類函數應採 Implicit Differentiation 。
2. 已知若 , 。若像上面的式子,我們將 或 均視為 這樣的替代變數 ( 事實上本來就是 變數項的替代函數 ) ,則微分方法其實是一樣的。
3 2 3y x xy y
ny u 1nd duy n u
dx dx
3y y nu y
x
26
3.8 Implicit Differentiation ( 隱微分 )
EX : 2 2 3y xy x
sol : 各別做
2 2[ ]' [3]'y xy x
2 2d dyy y
dx dx
[ ( )]d d d dyx y x y x y x y
dx dx dx dx
2 2dx x
dx
∴ 2 2[ ] 3dy xy x
dx
2 2 3dy dy dy x y xdx dx dx
2 2dy dyy x x ydx dx
(2 ) 2dy
y x x ydx
2
2
dy x y
dx y x
27
3.8 Implicit Differentiation ( 隱微分 )
EX :324 2 2xy y y , 求微分
dy
dx
sol : 4 4 4 3 4( ) 4d d d dyx y x y y x xy y
dx dx dx dx
d dyy
dx dx
3 12 2(2 ) 3
d dyy y
dx dx
1234 3 0
dy dy dyxy ydx dx dx
123[4 3 ] 0dy
xy ydx
123
1
4 3
dy
dx xy y
28
3.8 Implicit Differentiation ( 隱微分 )
EX : 求 的斜率 [ 或切線於 (1,3)]
sol :
2 23 12x y
2 2(3 ) 12d d d
x ydx dx dx
6 2 0dy
x ydx
2 6dyy xdx
3dy x
dx y
∴截點 (1,3) 的斜率為:3
13
dy
dx