If {x^y} = {e^{x - y}},then frac{dy}{dx} = | vedclass

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Question

(A) Given the equation ${x^y} = {e^{x - y}}$.Taking the natural logarithm on both sides,we get:$y \log x = x - y$Rearranging the terms to solve for $y

Vedclass · Accepted Answer

$\log x \cdot [\log (ex)]^{-2}$

Vedclass · Answer

(A) Given the equation ${x^y} = {e^{x - y}}$.Taking the natural logarithm on both sides,we get:$y \log x = x - y$Rearranging the terms to solve for $y$:$y(1 + \log x) = x$$y = \frac{x}{1 + \log x}$Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:$\frac{dy}{dx} = \frac{(1 + \log x)(1) - x(\frac{1}{x})}{(1 + \log x)^2}$$\frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2}$$\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}$Since $1 + \log x = \log e + \log x = \log (ex)$:$\frac{dy}{dx} = \log x \cdot [\log (ex)]^{-2}$.

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

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