A) \[\frac{2(1+\log x)}{{{(2+\log x)}^{2}}}\]
B) \[\frac{1+\log x}{{{(2+\log x)}^{2}}}\]
C) \[\frac{2}{2+\log x}\]
D) \[\frac{2(1-\log x)}{{{(2+\log x)}^{2}}}\]
E) \[\frac{2+\log x}{{{(2+\log x)}^{2}}}\]
Correct Answer: A
Solution :
We have\[{{x}^{y}}={{e}^{2(x-y)}}\] Taking log on both sides, we have \[\log ({{x}^{y}})=\log ({{e}^{2(x-y)}})\] \[\Rightarrow \] \[y\log x=2(x-y)\] \[\Rightarrow \] \[y=\frac{2x}{\log x+z}\] ?. (i) Differentiating w.r.t.\[x,\]we have \[y.\frac{1}{x}+\frac{dy}{dx}\log x=2\left( 1-\frac{dy}{dx} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}(\log x+2)=2-\frac{y}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2x-y}{x(2+\log x)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2x-\frac{2x}{2+\log x}}{x(2+\log x)}\][using Eq.(i)] \[=\frac{2x(1+\log x)}{x{{(2+\log x)}^{2}}}\] \[=\frac{2(1+\log x)}{{{(2+\log x)}^{2}}}\]
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