求导arctany/x=根号[ln(x^2+y^2) ] .根号在ln外面的

两边对x求导得1/[1+(y/x)^2]*(y/x)'=1/[ln(x^2+y^2) ] *[ln(x^2+y^2) ] '1/[1+(y/x)^2]*(y'x-y)/x^2=1/[2ln(x^2+y^2) ] *1/(x^2+y^2) *(x^2+y^2) '1/[1+(y/x)^2]*(y'x-y)/x^2=1/[2ln(x^2+y^2) ] *1/(x^2+y^2) *(2x+2yy') 1/[x^2+y^2]*(y'x-y)=1/[2ln(x^2+y^2) ] *1/(x^2+y^2) *(2x+2yy') (y'x-y)=1/[2ln(x^2+y^2) ] *(2x+2yy') 解出来y'即可