如图,
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如图,
如图,
如图,
两边对x求导:f1(-yz'x/z^2)+f2((xz'x-z)/x^2)=0 z'x=(zf2/x^2)/(f2/x-yf1/z^2)
两边对y求导:f1((z-yz'y/z^2)+f2(z'y/x)=0 z'y=(-f1/z)/(f2/x-yf1/z^2)
xz'x+yz'y=x[(zf2/x^2)/(f2/x-yf1/z^2)]+y[(-f1/z)/(f2/x-yf1/z^2)]
=[zf2/x-yf1/z]/[f2/x-yf1/z^2)]
=z
似曾相识 忘记了
几年不接触 想不起来偏导数怎么求了