花鸨皮丁儿
幼苗
共回答了14个问题采纳率:85.7% 举报
z^2-xy+z=1, 求 ∂²z/∂x∂y
等式两边对y求偏导,得 2zz'-x+z'=0,得 z'=x/(1+2z),
等式两边对x求偏导,得 2zz'-y+z'=0, 得 z'=y/(1+2z),
进而得 z'' = (1+2z-2yz')/(1+2z)^2 = [1+2z-2xy/(1+2z)]/(1+2z)^2
= [(1+2z)^2-2xy]/(1+2z)^3.
记 F=x+y-u-v=0, G=x/y-sinu/sinv=0, 其中 u=u(x,y),v=v(x,y).
则 F'=1-u'-v'=0, G'=1/y-(u'cosusinv-v'sinucosv)/(sinv)^2=0,
即 u'+v'=1
u'cosusinv-v'sinucosv=(sinv)^2/y
当 sin(u+v)≠0 时联立解得
u'= [(sinv)^2/y+sinucosv]/sin(u+v),
v'=[-(sinv)^2/y+cosusinv]/sin(u+v).
1年前
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