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If $x^y=e^{x-y}$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{\log x}{(1+\log x)^2}$
$x^y=e^{x-y}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x} \log x+y \cdot \frac{1}{x}=1-\frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}=\frac{x-y}{x(1+\log x)} \\ & \Rightarrow \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \quad\left[\text { Putting } y=\frac{x}{1+\log x} \text { from (i)] }\right.\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x} \log x+y \cdot \frac{1}{x}=1-\frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}=\frac{x-y}{x(1+\log x)} \\ & \Rightarrow \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2} \quad\left[\text { Putting } y=\frac{x}{1+\log x} \text { from (i)] }\right.\end{aligned}$
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