Re: [微分] 隐函数微分

楼主: Honor1984 (希望愿望成真)   2014-04-27 19:51:31
※ 引述《anger50322 (江)》之铭言:
: 欸欸用。我(x+y)/(x-y)=1/y,求dy/dx=?
: 答案是 (1-y)/(x+2y+1)
: 算不出这个答案
: http://i.imgur.com/TCtpjgh.jpg附上算式,请教各位
你算的跟答案算的其实是一样的
2y/(x - y) = (1 - y)/y
2x/(x - y) = (1 + y)/y
y' = 2y /[(x + y)^2 + 2x]
= (1 - y)(x - y)/{y [(x - y)^2/y^2 + 2x]}
= (1 - y)/{y [(x - y)/y^2 + 2x/(x - y)]}
= (1 - y)/[(x - y)/y + (1 + y)]
= (1 - y)/[(x + y) + (1 + y)]
= (1 - y)/[x + 2y + 1]

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