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Question

If xy=exy, then show that dydx=logx(1+logx)2

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Solution

Gicen, xy=exy
Taking log on both sides,
ylogx=xy ... (i)
y(1+logx)=x
y=x(1+logx) .... (ii)
Differentiating both sides of equation (i) using quotient rule, we get
dydx=(1+logx).1x.(1x)(1+logx)2
dydx=1+logx1(1+logx)2
dydx=logx(1+logx)2

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