y= tan^2 x
用y表示d^2y/dx^2
最好不要skip步驟，本人比較頓
記住用y表示，不是x
thx
ce程度微分，急（40分）
2008-03-24 8:49 pm
回答 (3)
2008-03-24 9:25 pm
✔ 最佳答案
dy/dx = 2 tanx sec^2 xd^2y/dx^2 = 2tanx (2secx tanx secx) + 2sec^2 x (sec^2 x)
= 4tan^2 x sec^2 x + 2sec^4 x
= 4tan^2 x (1+ tan^2 x ) + 2(1+ tan^2 x)^2
=4y(1+ y) + 2(1 +y)^2
=4y + 4 y^2 + 2 (1+ 2y +y^2)
=4y + 4 y^2 + 2+ 4y +2 y^2
=6 y^2 + 8y+ 2
參考: my math knowledge
2008-03-24 9:36 pm
dy/dx = (2tanx) (sec^2 x)
= 2tanx sec^2 x
d^2y/dx^2 = 2 [(tanx)(2secx)(secx tanx) + (sec^2 x)(sec^2 x)]
= 2 (tan^2 x sec^2 x + sec^4 x)
= 2sec^2 x (tan^2 x + sec^2 x)
= 2sec^2 x (y + sec^2 x) .........Sub. y= tan^2 x into the equation......
d^2y/dx^2 / 2sec^2 x = y + sec^2 x
y = d^2y/dx^2 / 2sec^2 x - sec^2 x
= 2tanx sec^2 x
d^2y/dx^2 = 2 [(tanx)(2secx)(secx tanx) + (sec^2 x)(sec^2 x)]
= 2 (tan^2 x sec^2 x + sec^4 x)
= 2sec^2 x (tan^2 x + sec^2 x)
= 2sec^2 x (y + sec^2 x) .........Sub. y= tan^2 x into the equation......
d^2y/dx^2 / 2sec^2 x = y + sec^2 x
y = d^2y/dx^2 / 2sec^2 x - sec^2 x
2008-03-24 9:36 pm
dy/dx = 2 tanx sec^2 x
d^2y/dx^2 = 2tanx (2secx tanx secx) + 2sec^2 x (sec^2 x)
= 4tan^2 x sec^2 x + 2sec^4 x
= 4tan^2 x (1+ tan^2 x ) + 2(1+ tan^2 x)^2
=4y(1+ y) + 2(1 +y)^2
=4y + 4 y^2 + 2 (1+ 2y +y^2)
=4y + 4 y^2 + 2+ 4y +2 y^2
=6 y^2 + 8y+ 2
d^2y/dx^2 = 2tanx (2secx tanx secx) + 2sec^2 x (sec^2 x)
= 4tan^2 x sec^2 x + 2sec^4 x
= 4tan^2 x (1+ tan^2 x ) + 2(1+ tan^2 x)^2
=4y(1+ y) + 2(1 +y)^2
=4y + 4 y^2 + 2 (1+ 2y +y^2)
=4y + 4 y^2 + 2+ 4y +2 y^2
=6 y^2 + 8y+ 2
參考: dad & mum
收錄日期: 2021-04-29 21:46:32
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https://hk.answers.yahoo.com/question/index?qid=20080324000051KK01119