If x > 0 , x y = e x − y , then d x d y is equal to, ( 1 + l o g x ) 2 1 , ( 1 + l o g x ) 2 l o g x , ( 1 + l o g x l o g x ) 2, 1 + l o g x ( l o g x ) 2

If x > 0 , x y = e x − y , then d x d y is equal to

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If $x>0, x^y=e^{x-y}$, then $\frac{d y}{d x}$ is equal to

MathematicsDifferentiationAP EAMCETAP EAMCET 2004

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The correct answer is: $\frac{\log x}{(1+\log x)^2}$

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