已知sin(xy)-ln((x+1)/y)=1 求dy/dx

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已知sin(xy)-ln((x+1)/y)=1 求dy/dx
已知sin(xy)-ln((x+1)/y)=1 求dy/dx

已知sin(xy)-ln((x+1)/y)=1 求dy/dx
cos(xy)(xy)'-[y/(x+1)-y']/y^2=0
cos(xy)(y+xy')-[y/(x+1)-y']/y^2=0
y'[xcos(xy)+1/y^2]=1/y(x+1)-ycos(xy)
y'=[1/y(x+1)-ycos(xy)]/[xcos(xy)+1/y^2]

d(sin(xy)-ln((x+1)/y))=d(1)=0
d(sin(xy)-ln((x+1)/y))=cos(xy)*d(xy)-y/(x+1)*d((x+1)/y)
=cos(xy)*(x*dy+y*dx)-y/(x+1)*(dx/y-(x+1)*dy/y^2)
=0
于是 (cos(xy)*y-(y/(x+1))*(1/y))*dx=(-cos(xy)*x-(x+1)/y^2)*dy
(cos(xy)*y-1/(x+1))/(-cos(xy)*x-(x+1)/y^2)=dy/dx