If $x^y=e^{x-y}$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

$x^y=e^{x-y}$ $\begin{aligned} & \Rightarrow \frac{d y}{d x} \log x+y \cdot \frac{1}{x}=1-\frac{d y}{d x} \ & \Rightarrow \frac{d y}{d x}=\frac{x-y}{x(1+\log x)} \ & \Rightarrow \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \quad\left[	ext { Putting } y=\frac{x}{1+\log x} 	ext { from (i)] }ight.\end{aligned}$

If $x^y=e^{x-y}$, then $\frac{d y}{d x}=$

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