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Question

If x1+y+y1+x=0, then dydx is equal to

A
1(1+x)2
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B
1(1+x)2
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C
1(1x)2
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D
None of these
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Solution

The correct option is B 1(1+x)2
Given x1+y+y1+x=0
x1+y=y1+x
1+yy=1+xx
(1+yy)2=(1+xx)2
1+yy2=1+xx2
x2+x2y=y2+y2x
x2y2=y2xx2y
(x+y)(xy)=xy(yx)
x+y+xy=0y=x1+x (xy)
Differentiate on both sides w.r.t x
1+dydx+xdydx+y=0
dydx=(1+y)1+x
=(1x1+x)1+x
=1(1+x)2
dydx=1(1+x)2

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