求dy/dx, x^2y^2-4y=1这题怎么做?

已知x=y^2-4y,求dy/dx 求dy/dx, x^2y^2-4y=1这题怎么做? x^y-y^x=2,求dy/dx 求dy/dx=(x-y+5)/(x+y-2) 设x^2+y^2=4,求dy/dx 设y=√(x^2-1)求dy/dx x^2+xy+y^3=1,求dy/dx 求dy/dx-2y/x=1的通解 设y=sin2x/(1+x^2),求dy/dx, 设y=sin2x/(1+x^2),求dy/dx, 求dy/dx y=e^(2x+1) 求方程(2x+y-4)dx+(x+y-1)dy=0的通解. x^2+y^2=e^y求dy/dx 求dy/dx=x/y+(cosx/y)^2通解 (x^2)dy+（y^2)dx=dx-dy y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy dy/dx+y/3=1/3(1-2x)y^4求通解 由方程y^2*sinx+e^y+2x=1,求dx/dy是dy/dx 打错了