If x^{y}=e^{x-y} then frac{d y}{d x} is e | vedclass

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

Vedclass · Accepted Answer

$ \frac{\log x}{(1+\log x)^{2}} $

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(A) Given that,$ x^{y}=e^{x-y} $.Taking the natural logarithm on both sides,we get:$ y \log x = x - y $Rearranging the terms to isolate $ y $:$ y + y \log x = x $$ y(1 + \log x) = x $$ y = \frac{x}{1 + \log x} $Now,differentiating with respect to $ x $ using the quotient rule $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} $:$ \frac{dy}{dx} = \frac{(1 + \log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \log x)}{(1 + \log x)^2} $$ \frac{dy}{dx} = \frac{(1 + \log x)(1) - x \cdot (\frac{1}{x})}{(1 + \log x)^2} $$ \frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} $$ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} $

If $ x^{y}=e^{x-y} $ then $ \frac{d y}{d x} $ is equal to

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