20點(Amaths/PureMaths高手請進)D arc

👨🏻‍🏫 學術台數學

[匿名]

2011-04-06 3:26 am

請問點樣計
Differential arcsin√1-x2
x2 即係 x既2次方...

唔該連埋 steps + answer
唔該晒高手 ~

更新1:

唔好意思打溜左...條問題係要你用 chain rule... 問題係 Apply the chain rule to find the differential arcsin√1-x2

更新2:

請問你地兩個係唔係用緊 Chain rule 計? 問題要我用 chain rule 計 Apply the chain rule to find the differential arcsin√1-x2

更新3:

兩位師兄, 介唔介意用小畫家 / Microsoft Word 做一次 send 去我 email 到? 比較難睇係依到.../_\ 唔該晒...thanks... [email protected]

回答 (2)

myisland8132

2011-04-06 4:37 am

✔ 最佳答案

(arcsin x)' = 1/√(1 - x^2)

[arcsin √(1 - x^2)]'

= {1/√(1 - [√(1 - x^2)]^2)} [√(1 - x^2)]'

= (1/x)[-x/√(1 - x^2)]

= -1/√(1 - x^2)

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匿名

2011-04-06 4:12 am

y = sin^-1 √(1-x^2)
siny = √(1-x^2)/1
cosy = x/1 = x
secy = 1/x
siny = √(1-x^2) ---------(**)
Differentiate both sides of (**),
cosy * dy/dx = (1/2)(1-x^2)^(-1/2) (-2x) = -x/√(1-x^2)
dy/dx = -x/√(1-x^2) * sec y
dy/dx = -1/√(1-x^2)

2011-04-05 21:22:44 補充：
問題要我用 chain rule 計 << 我絕對有用
siny = √(1-x^2)
dsiny/dy * dy/dx = d√(1-x^2) /d(1-x^2) * d(1-x^2)/dx

參考: Hope the solution can help you^^”

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