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  4. If xy=e(x-y), then find dy/dx

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Q. If $ {{x}^{y}}={{e}^{(x-y)}}, $ then find $ dy/dx $

J & K CETJ & K CET 2014

Solution:

We have, $ {{x}^{y}}={{e}^{(x-y)}} $ Taking log on both sides, we get $ y\log x=(x-y)\log \,e $
$ \Rightarrow $ $ y\log x=x-y $ ?.(i) On differentiating w. r. t. x, we get $ y.\frac{1}{x}+\log x.\frac{dy}{dx}=1-\frac{dy}{dx} $
$ \Rightarrow $ $ \frac{dy}{dx}(\log x+1)=1-\frac{y}{x} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{x-y}{(\log \,x+1)x} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{x-\frac{x}{\log x+1}}{x(\log x+1)} $ $ \left( from\,Eq.\,(i),\,\,y=\frac{x}{\log x+1} \right) $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{x\log x+x-x}{x{{(\log \,x+1)}^{2}}} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{x\log x}{x{{(\log x+1)}^{2}}} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}} $