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Question

If f(x,y)=1x2+1xy+logxlogyx2+y2, then xfx+yfy is equal to

A
0
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B
f
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C
2f
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D
2f
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Solution

The correct option is D 2f
Given,
f(x,y)=1x2+1xy+logxlogyx2+y2
So, by partially differentiating the function w.r.t x and multiplying with x, we get,
xfx=2x21xy+(x2+y2)(logxlogy)(2x2)(x2+y2)2
similiarly, if we partially differentiate w.r.t y and multiply it with y, we get,
yfy=1xy+(x2+y2)(logxlogy)(2y2)(x2+y2)2
Adding both the equations we get,
xfx+yfy=2x22xy(logxlogy)(2x2+2y2)(x2+y2)2=2x22xy2(logxlogy)(x2+y2)2=2f

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