If xloge(logex) $-$ x2 + y2 = 4(y > 0), then ${{dy} \over {dx}}$ at x = e is equal to :

Question

If xlog_e(log_ex) $-$ x² + y² = 4(y > 0), then ${{dy} \over {dx}}$ at x = e is equal to :

${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$

${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$

${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$

${e \over {\sqrt {4 + {e^2}} }}$

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