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Question

If ln⟮(e−1)exy+x2⟯=x2+y2, then ⟮dydx⟯(1,0) is given by

A
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B
2
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C
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D
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Solution

The correct option is B 2
((e1)exy+x2)=e(x2+y2)
(e1)exy(y+xdydx)+exy(0)+2x
=e(x2+y2)(2x+2ydydx)
(e1)dydx+2=e(2+0)
dydx=2e2e1=2

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