Question

求由方程cos（xy）=x^2*y^2 所确定的y的微分

Answer 1

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy

-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy

-2x^2*ydy-sin(xy)xdy=2xy^2*dx+sin(xy)ydx

-[2x^2*y+sin(xy)x]dy=[2xy^2+sin(xy)y]dx

dy/dx=-[2xy^2+sin(xy)y]/[2x^2*y+sin(xy)x]