e^(x-y^2) 求导

e^(x-y^2)=0 这是隐式方程求导：[e^(x-y^2)]'=0 →e^(x-y^2)·(1-2y·y')=0 ∵e^(x-y^2)＞0,∴1-2y·y'=0.则y'=1/(2y).e^(x-y^2)=0 则x-y^2=1.则y=√(x-1) 则y'=1/(2y)=1/(2√(x-1) )