设y=y(x)是由方程ln(x^2+y^2)=arctan(y/x)+ln2-π/4确定的隐函数,求dy|(1,1)
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对方程两边同时微分,得:
d[ln(x^2+y^2)
=(2xdx+2ydy)/(x^2+y^2)
d[arctan(y/x)+ln2-π/4]
=(xdy-ydx)/(x^2-y^2),
——》(2xdx+2ydy)/(x^2+y^2)=(xdy-ydx)/(x^2-y^2)
——》dy/dx=(2x^3+y^3-2xy^2+x^2y)/(x^3+2y^3+xy^2-2x^2y)=1.