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求由方程cos(xy)=x^2*y^2 所确定的y的微分
人气:429 ℃ 时间:2019-09-20 13:16:13
解答
-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]dxdy/dx=-[2xy^2+sin(xy)y]/[2x^2*y+sin(xy)x]
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