A) \[\log x/{{(1+\log x)}^{2}}\]
B) \[(\log x)\,/\,(1+\log x)\]
C) \[(1-\log x)\,/\,(1+\log x)\]
D) \[(\log x)\,/\,{{(1-\log x)}^{2}}\]
Correct Answer: A
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 \[y.\frac{1}{x}+\log x.\frac{dy}{dx}=1-\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}(\log x+1)=1-\frac{y}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x-y}{(\log \,x+1)x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x-\frac{x}{\log x+1}}{x(\log x+1)}\] \[\left( from\,Eq.\,(i),\,\,y=\frac{x}{\log x+1} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x\log x+x-x}{x{{(\log \,x+1)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{x\log x}{x{{(\log x+1)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\log x}{{{(1+\log x)}^{2}}}\]
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