f'x=1+(y-1)/[y*√(1-x/y)]
f'y=arcsin(x/y)^1/2-x(y-1)/[y^2*√(1-x/y)]
對(duì)x偏導(dǎo)將y看做常數(shù)
對(duì)y偏導(dǎo)將x看作常數(shù)就行.不是復(fù)合函數(shù)嗎···怎么能直接導(dǎo)這是二元函數(shù)啊，當(dāng)然是求偏導(dǎo)啊。arcsin(x/y)^1/2····能直接導(dǎo)成1/√(1-x/y) ????復(fù)合函數(shù)不能直接導(dǎo)的啊復(fù)合函數(shù)f(g(x))'=f'(g(x))*g'(x)所以求導(dǎo)后還要對(duì)內(nèi)層函數(shù)求導(dǎo)啊arcsin(x/y)^1/2····當(dāng)然能直接導(dǎo)成1/√(1-x/y) 還要乘以(x/y)^1/2的導(dǎo)數(shù)啊。。