求dy/dx, x^2y^2-4y=1这题怎么做?
问题描述:
求dy/dx, x^2y^2-4y=1这题怎么做?
答
x²y²-4y=1,2xy²+2x²y(dy/dx)-4(dy/dx)=0,dy/dx=(xy²)/(2-x²y)
求dy/dx, x^2y^2-4y=1这题怎么做?
x²y²-4y=1,2xy²+2x²y(dy/dx)-4(dy/dx)=0,dy/dx=(xy²)/(2-x²y)