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Question

If xx+yx=1, prove that dydx=-xx 1+log x+yx·log yx·yx-1

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Solution

We have, xx+yx=1elogxx+elogyx=1ex logx+ex logy=1
Differentiating with respect to x using chain rule,
ddxexlogx+ddxex logy=ddx1ex logxddxx logx+ex logyddxx logy=0ex logxxddxlogx+logxddxx+elogyxxddxlogy+logyddxx=0xxx1x+logx1+yxx1ydydx+logy1=0xx1+logx+yxxydydx+logy=0yx×xydydx=-xx1+logx+yxlogyxyx-1dydx=-xx1+logx+yxlogydydx=-xx1+logx+yxlogyxyx-1

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