求(1+y^2)dx=(arctany-x)dy的通解.

求(1+y^2)dx=(arctany-x)dy的通解. 设y=2arctany/x,求dy/dx,dy^2/d^2x. 设函数y=y(x)由方程ln(x^2+y^2)^1/2=arctany/x所确定,求dy/dx. x=y+arctany 求隐函数的导数dy/dx. ln(x²+y²)½=arctany/x 求dy/dx 1 求方程（1+y^2）dx=(arctany - x)dy的通解 2求方程（x-2xy- y^2）y’+ =0的通解第二题不完整 补充下 求方程（x-2xy-y^2）y'+y^2=0的通解 ln[根号（x^2+y^2）] =arctany/x 求dy1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x-y)则dy=(x+y)/(x-y)*dx请问两边对求导是怎么得出1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+( ln[根号（x^2+y^2）] =arctany/x 求dy 求隐函数的导数 arctany/x=ln根号下（x^2+y^2) 设z=(x^2+y^2)*e^(-arctany/x) ,求dz f(x,y)=x平方arctany求微分 急 x+y=arctany 隐函数求导数 y=x+arctany微分 求隐函数的导数?1 y=x+ 1/2iny2 2^x+2y=2^x+y3 x=y+arctany 已知xy+x+y=1,则arctanx+arctany等于 求下列方程所确定的函数y=y（x）的二阶导数：y=1+xe^y（1）y=1+xe^y（2）y=x+arctany 已知x+arctany=y,求函数y=(x)的导数y'? 求隐函数的导数y'.y=cos(x+y) y=x+lny/2 x=y+arctany 这三个