A) \[\frac{(1+2e)}{\sqrt{4+{{e}^{2}}}}\]
B) \[\frac{(1+2e)}{2\sqrt{4+{{e}^{2}}}}\]
C) \[\frac{e}{\sqrt{4+{{e}^{2}}}}\]
D) \[\frac{(2e-1)}{2\sqrt{4+{{e}^{2}}}}\]
Correct Answer: D
Solution :
Given, \[x{{\log }_{e}}(lo{{g}_{e}}x)-{{x}^{2}}+{{y}^{2}}=4\] ...(i) Differentiating both sides of (i) w.r.t x, we get \[x\frac{1}{{{\log }_{e}}x}.\frac{1}{x}+{{\log }_{e}}(lo{{g}_{e}}x)-2x+2y.\frac{dy}{dx}=0\] At\[x=e,1-2e+2y.\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{2e-1}{2y}\]?(ii) Also, from (i), \[-{{e}^{2}}+{{y}^{2}}=4\]\[\Rightarrow \]\[y=\sqrt{4+{{e}^{2}}}\] \[\therefore \]From (n),\[\frac{dy}{dx}=\frac{2e-1}{2\sqrt{4+{{e}^{2}}}}\]
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