Find (d^2 y) / dx^2 in terms of x and y of the following plane curves:
1. tan y = loge x^2 (e is the base)
2. y = x e^y
(must use the method ln both sides.)
更新1:
六呎將軍, u need to differentiate twice, not just once..
更新2:
help me
Urgent question
2010-05-01 8:32 pm
Show the steps very clearly.
Find (d^2 y) / dx^2 in terms of x and y of the following plane curves:
1. tan y = loge x^2 (e is the base)
2. y = x e^y
(must use the method ln both sides.)
Find (d^2 y) / dx^2 in terms of x and y of the following plane curves:
1. tan y = loge x^2 (e is the base)
2. y = x e^y
(must use the method ln both sides.)
回答 (2)
2010-05-01 9:10 pm
✔ 最佳答案
1)圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May10/Crazydiff1.jpg
2)
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May10/Crazydiff2.jpg
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參考: My Maths knowledge
2010-05-01 9:16 pm
2. y = xe^y
lny = ln(xe^y) = lnx + lne^y = lnx + y
Differentiate both sides w.r.t. x
1/y dy/dx = 1/x + dy/dx
(1/y - 1)dy/dx = 1/x
dy/dx = y/[x(1 - y)]
2010-05-01 13:16:17 補充：
Differentiate both sides w.r.t. x
d^2y/dx^2 = {x(1 - y)dy/dx - y[x(-dy/dx) + (1 - y)]}/[x(1 - y)]^2
= [y + y^2/(1 - y) - y + y^2] / [x(1 - y)]^2
= (2y^2 - y^3)/[x^2(1 - y)^3]
lny = ln(xe^y) = lnx + lne^y = lnx + y
Differentiate both sides w.r.t. x
1/y dy/dx = 1/x + dy/dx
(1/y - 1)dy/dx = 1/x
dy/dx = y/[x(1 - y)]
2010-05-01 13:16:17 補充：
Differentiate both sides w.r.t. x
d^2y/dx^2 = {x(1 - y)dy/dx - y[x(-dy/dx) + (1 - y)]}/[x(1 - y)]^2
= [y + y^2/(1 - y) - y + y^2] / [x(1 - y)]^2
= (2y^2 - y^3)/[x^2(1 - y)^3]
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