已知方程ln√(x^2-Y^2)=arccot(y/x)求dy/dxdy/dx=x+y/x-y

Question

已知方程ln√(x^2-Y^2)=arccot (y/x) 求dy/dx
dy / dx = x+y / x-y

Answer 1

已知方程ln√(x²-Y²)=arccot (y/x) ,求dy/dx
F(x,y)=ln√(x²-Y²)-arccot (y/x)=(1/2)ln(x²-y²)-arccot(y/x)≡0
dy/dx=-(∂F/∂x)/(∂F/∂y).(1)
其中∂F/∂x=(1/2)[2x/(x²-y²)]+[-(y/x²)/(1+y²/x²)]=x/(x²-y²)-y/(x²+y²)
∂F/∂y=(1/2)[-2y/(x²-y²)]+[(1/x)/(1+y²/x²)]=-y/(x²-y²)+x/(x²+y²)
代入(1)式即得:
dy/dx=-[x/(x²-y²)-y/(x²+y²)]/[-y/(x²-y²)+x/(x²+y²)]=-[x(x²+y²)-y(x²-y²)]/[-y(x²+y²)+x(x²-y²)]
=-(x³+xy²-yx²+y³)/(-yx²-y³+x³-xy²)=(x³+xy²-yx²+y³)/(y³+xy²+yx²-x³)