xy=y(−x)
Taking logerithm with respect to base e on both sides
lnxy=lny(−x)
=> ylnx=−xlny
Differentiating with respect to x
dxd(ylnx)=−dxd(xlny)
=> ydxd(lnx)+lnxdxdy=−xdxd(lny)−lnydxd(x)
=> y.x1+lnxdxdy=−x.y1.dxdy−lny
=> lnxdxdy+x.y1.dxdy = −xy−lny
=> x(ylnx+x)dxdy=−y(y+xlny)
=>
=> dxdy= −x(ylnx+x)y(y+xlny)
So , y' = −x(ylnx+x)y(y+xlny)
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