arctan(x/y)=ln(√（x^2+y^2)求d^2y/dx^2

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已知arctan(x/y)=ln[√(x²+y²)],求d²y/dx².
F(x,y)=arctan(x/y)-ln[√(x²+y²)]=0
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[(1/y)/(1+x²/y²)-x/(x²+y²)]/[(-x/y²)/(1+x²/y²)-y/(x²+y²)]
=[(y-x)/(x²+y²)]/[(x+y)/(x²+y²)]=(y-x)/(x+y)
d²y/dx²=[(x+y)(y′-1)-(y-x)(1+y′)]/(x+y)²=(2xy′-2y)/(x+y)²=[2x(y-x)/(x+y)-2y]/(x+y)²
=-2(x²-y²)/(x+y)²=-2(x-y)/(x+y)=2(y-x)/(y+x)

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