求由方程xsiny-ycosx=2确定的隐函数y=yx的导数dy/dx

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求由方程xsiny-ycosx=2确定的隐函数y=yx的导数dy/dx

对两边求导.
xcosy×dy/dx+siny-dy/dx×cosx+ysinx=0
(xcosy-cosx)dy/dx=-siny-ysinx
所以dy/dx=(-siny-ysinx)/ ( xcosy-cosx ) .

(y×sinX-siny)/(xcosy-cosx)