dy=(1+y)/[1+(x+y)^2]dx+(1+x)/[1+(x+y)^2]dy
所以
dy/dx=(1+y)/[(x+y)^2-x]
y=tan(x+y)
y'=[tan(x+y)]'
=sec^2(x+y)*(x+y)'
=sec^2(x+y)*(1+y')
y'=sec^2(x+y)/[1-sec^2(x+y)]
=sec^2(x+y)/(-tan^2(x+y))
=-1/sin^2(x+y)
=-csc^2(x+y).
两边分别对x求导
y'=1/1+(x+y)^2 * (1+y')
化简整理就得到答案了，祝你成功！
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