If $x^y=e^{x-y}$,prove that $\Large \frac{dy}{dx}=\frac{log x}{(1+log x)^2}$

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$x^y=e^{x-y}$
Take log on both sides
$y\; log x=(x-y) log\; e \qquad (But\; log\;e=1)$
=>$y\; log\; x=x-y \qquad =>x=y(1+log\; x)$
differentiate on both sides,
$y.\large\frac{1}{x}+log\; x .\frac{dy}{dx}=1-\frac{dy}{dx}$
$=>\large\frac{dy}{dx}(1+\log x)=1-\frac{y}{x}$
Therefore $\large\frac{dy}{dx}=\frac{x-y}{x(1+log\;x)}$
Substituting for x-y, and x we get
$\large\frac{dy}{dx}=\frac{y\; log\; x}{y(1+log\;x)^2}$
Therefore $\large\frac{dy}{dx}=\frac{log \;x}{(1+\log\; x)^2}$
answered Mar 23, 2013 by