y = √[x-√(x²-1)]
更新1:
[sqrt(x^2-1) - x]/{2sqrt[x-sqrt(x^2-1)]sqrt(x^2-1)}如何計算到 -sqrt[x - sqrt(x^2-1)]/2sqrt(x^2-1)的? 我計到 [sqrt(x^2-1) - x]/{2sqrt[x-sqrt(x^2-1)]sqrt(x^2-1)} 但不懂出到-sqrt[x - sqrt(x^2-1)]/2sqrt(x^2-1)這個答案
中4附加數問題 求導數
回答 (3)
2008-08-06 4:48 pm
✔ 最佳答案
dy/dx = d{sqrt[x - sqrt(x^2 - 1)]}/d[x - sqrt(x^2 - 1)] x d[x - sqrt(x^2-1)]/dx= 1/{2sqrt[x - sqrt(x^2-1)]} x {dx/dx - d[sqrt(x^2-1)]/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - d[sqrt(x^2-1)]/d(x^2-1) x d(x^2-1)/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x { 1 - 1/[2sqrt(x^2-1)] x (2x)}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - x/sqrt(x^2-1)}
= [1-x/sqrt(x^2-1)]/{2sqrt[x-sqrt(x^2-1)]}
= [sqrt(x^2-1) - x]/{2sqrt[x-sqrt(x^2-1)]sqrt(x^2-1)}
= -sqrt[x - sqrt(x^2-1)]/2sqrt(x^2-1).
2008-08-06 12:32:04 補充：
Please note that A/sqrtA = sqrtA. So [sqrt(x^2-1) - x]/sqrt[x-sqrt(x^2-1)] = -[x-sqrt(x^2-1)]/sqrt[x-sqrt(x^2-1)] = -sqrt[x-sqrt(x^2-1)].
2008-08-06 7:12 pm
dy/dx = d{sqrt[x - sqrt(x^2 - 1)]}/d[x - sqrt(x^2 - 1)] x d[x - sqrt(x^2-1)]/dx
= 1/{2sqrt[x - sqrt(x^2-1)]} x {dx/dx - d[sqrt(x^2-1)]/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - d[sqrt(x^2-1)]/d(x^2-1) x d(x^2-1)/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x { 1 - 1/[2sqrt(x^2-1)] x (2x)}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - x/sqrt(x^2-1)}
= [1-x/sqrt(x^2-1)]/{2sqrt[x-sqrt(x^2-1)]}
= [sqrt(x^2-1) - x]/{2sqrt[x-sqrt(x^2-1)]sqrt(x^2-1)}
= -sqrt[x - sqrt(x^2-1)]/2sqrt(x^2-1).
= 1/{2sqrt[x - sqrt(x^2-1)]} x {dx/dx - d[sqrt(x^2-1)]/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - d[sqrt(x^2-1)]/d(x^2-1) x d(x^2-1)/dx}
= 1/{2sqrt[x-sqrt(x^2-1)]} x { 1 - 1/[2sqrt(x^2-1)] x (2x)}
= 1/{2sqrt[x-sqrt(x^2-1)]} x {1 - x/sqrt(x^2-1)}
= [1-x/sqrt(x^2-1)]/{2sqrt[x-sqrt(x^2-1)]}
= [sqrt(x^2-1) - x]/{2sqrt[x-sqrt(x^2-1)]sqrt(x^2-1)}
= -sqrt[x - sqrt(x^2-1)]/2sqrt(x^2-1).
參考: me
2008-08-06 10:46 am
Using chain rule repeatedly, you should get
dy/dx = -√[x-√(x²-1)] / 2√(x²-1)
dy/dx = -√[x-√(x²-1)] / 2√(x²-1)
收錄日期: 2021-04-25 22:35:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080806000051KK00046