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Given the equation $x^{y}=e^{x-y}$. Taking the natural logarithm (ln) on both sides, we get: $$ln(x^{y}) = ln(e^{x-y})$$ Using the logarithm property $ln(a^b) = b \cdot ln(a)$, we have: $$y \cdot ln(x) = (x-y) \cdot ln(e)$$ Since $ln(e) = 1$, the equation simplifies to: $$y \cdot ln(x) = x - y$$
Rearrange the equation to isolate $y$: $$y \cdot ln(x) + y = x$$ $$y(ln(x) + 1) = x$$ $$y = \frac{x}{1 + ln(x)}$$
Now, differentiate $y$ with respect to $x$ using the quotient rule, which states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = x$ and $v = 1 + ln(x)$. So, $\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \frac{1}{x}$. Applying the quotient rule: $$\frac{dy}{dx} = \frac{(1 + ln(x))(1) - x(\frac{1}{x})}{(1 + ln(x))^2}$$ $$\frac{dy}{dx} = \frac{1 + ln(x) - 1}{(1 + ln(x))^2}$$ $$\frac{dy}{dx} = \frac{ln(x)}{(1 + ln(x))^2}$$
Thus, we have shown that: $$\frac{dy}{dx} = \frac{ln(x)}{(1 + ln(x))^2}$$
Final Answer: $\frac{log~x}{(1+log~x)^{2}}$
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