dororo988

求x導數
1)y=3-2√(x)
2)y=4√(x)-1

firoazen

1
-1/√(x)

2
2/√(x)

dororo988

我想要過程

manager

for q(1)
y  = 3 - 2 sqrt x = 3 - 2 x^(1/2)
y' = 0 - 2(1/2) x^(-1/2) = -1 / sqrt x

for q(2)
y = 4 sqrt x - 1 = 4 x^(1/2) - 1
y' = 4(1/2) x^(-1/2) - 0 = 2 / sqrt x

[ 本帖最後由 manager 於 2011-8-6 08:53 PM 編輯 ]

thomas3253

引用:

原帖由 dororo988 於 2011-8-6 07:12 PM 發表
求x導數
1)y=3-2√(x)
2)y=4√(x)-1

1)
y=3-2x^0.5
y'=(3-2x^0.5)'
y'=(3)'-(2x^0.5)'
y'=0-2(x^0.5)'
y=-2(0.5)[x^(0.5-1)]
y'=-1(x^-0.5)

2)
y=4x^0.5-1
y'=(4x^0.5-1)'
y'=(4x^0.5)'-(1)'
y'=4(x^0.5)'-0
y'=4(0.5)[x^(0.5-1)]
y'=2x^-0.5

桃子

y=3-2x^0.5
dy/dx=-2(0.5)x^-0.5
=-1/sqrt(x)

y=4(x^0.5)-1
dy/dx=-4(0.5)x^-0.5
=2/sqrt(x)

dororo988

但是我平時唔係這做
好似係用[f(x+△x)-f(x)]/△x
可否用這做
否則我跟你
但我唔識你的method

manager

引用:

原帖由 dororo988 於 2011-8-6 10:21 PM 發表
但是我平時唔係這做
好似係用[f(x+△x)-f(x)]/△x
可否用這做
否則我跟你
但我唔識你的method

You are mentioning the derivative from the first principle.

You can just differentiate and get the result unless you see the wordings "By the First Principle ..."
The above solutions involves this frequently used rule:

If then where r is any real number.

[ 本帖最後由 manager 於 2011-8-6 11:15 PM 編輯 ]

dororo988

題目寫
[從基本原理求下列函數對x的導數]
這算唔算叫我用第1條
若是可否用這公式help me做一次

[ 本帖最後由 dororo988 於 2011-8-7 04:27 PM 編輯 ]

夏娜醬大好

1.
dy/dx
= lim {[3-2√(x+h)]-[3-2√(x)]}/h
= 2 * lim [√x-√(x+h)]/h
= 2 * lim [√x-√(x+h)]/h * [√x+√(x+h)]/[√x+√(x+h)]
= 2 * lim (-1)/[√x+√(x+h)]
= 2 * (-1)/(2√x)
= -1/√x

2.
dy/dx
= lim {[4√(x+h)-1]-[4√(x)-1]}/h
= 4 * lim [√(x+h)-√(x)]/h
= 4 * lim [√(x+h)-√(x)]/h * [√(x+h)+√(x)]/[√(x+h)+√(x)]
= 4 * lim 1/[√(x+h)+√(x)]
= 4 * 1/(2√x)
= 2√x

dororo988

2 * lim (-1)/[√x+√(x+h)]
點解你知係-1

夏娜醬大好

[x-(x+h)]/h

dororo988

仲有一題
y=-2x^(3)+1

夏娜醬大好

lim { [-2(x+h)^3+1]-[-2x^3+1] }/h
= -2 * lim [ (x+h)^3-x^3 ]/h
= -2 * lim 3x^2+3xh+h^2
= -6x^2

dororo988

(a)^3-(b)^3
唔係=(a-b)(a^2+ab+b^2)?
唔係好明第2step點做到第3個step

SORRY 仲有一題想問
y=1/5x

[ 本帖最後由 dororo988 於 2011-8-7 08:05 PM 編輯 ]