if e to the power x + e to the power y = e to the power x + y then prove; dy/dx + e to the power x-y 0

2.if xy.log(x + y)=1

then find dy/dx

Given , ex+ey=ex+y dividing by ex+y  we get, exex+y +eyex+y =1ex-x-y+ey-x-y=1e-y+e-x=1on differentiating we get, ddxe-y+ddxe-x=ddx1-e-ydydx-e-x=0dydx=-e-xe-y=-ey-xor dydx+ey-x=0Hence proved.2)Given xylogx+y=1ylogx+y=1x=x-1On differentiating we get, yddxlogx+y+logx+ydydx=-1x2yx+y1+dydx+logx+ydydx=-1x2dydxyx+y+logx+y=-1x2-yx+ydydx=-1x2-yx+yyx+y+logx+yYou can further simplify for your practice.  

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