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If ln((e-1)e^(xy) +x^2)=x^2+y^2 then ((d...

If `ln((e-1)e^(xy) +x^2)=x^2+y^2` then `((dy)/(dx))_(1,0)` is equal to

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`"We have, "(e-1)e^(xy)+x^(2)=e^(x^(2))+y^(2)`
Differentiating both sides w.r.t. x, we get
`(e-1)cdote^(xy)cdot(xcdot(dy)/(dx)+y)+2x=e^(x^(2)+y^(2)).(2x+2y(dy)/(dx))`
Putting x=1 and y = 0, we get
`(e-1)((dy)/(dx))+2=e(2+0)`
`rArr" "((dy)/(dx))_((1,0))=2`
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