设In√x2+y2=arctan(y/x)，求dy/dx．
方程两边分别对x求导：(1/2)ln(x2+y2)=arctan(y/x)⇒(1/2)•[(2x+2y)•y′]/(x2+y2)=[(y′•x-y)/x2]/[1+(y/x)2]⇒(x+yy′)/(x2+y2)=(xy′-y)/(x2+y2)即x+yy′，=xy′，-y，经整理得dy/dx=y′=(x+y)/(x-y)．
设In√x2+y2=arctan(y/x)，求dy/dx．
方程两边分别对x求导：(1/2)ln(x2+y2)=arctan(y/x)⇒(1/2)•[(2x+2y)•y′]/(x2+y2)=[(y′•x-y)/x2]/[1+(y/x)2]⇒(x+yy′)/(x2+y2)=(xy′-y)/(x2+y2)即x+yy′，=xy′，-y，经整理得dy/dx=y′=(x+y)/(x-y)．
