If x loge (loge x) – x2 + y2 = 4 (y > 0), then dy/dx at x = e is equal to:

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

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

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

\(\frac{e}{{\sqrt {4 + {e^2}} }}\)

If x loge (loge x) – x2 + y2 = 4 (y > 0), then dy/dx at x = e is eq

1.	If x loge (loge x) – x2 + y2 = 4 (y > 0), then dy/dx at x = e is equal to:
A.	\(\frac{{\left( {1 + 2e} \right)}}{{2\sqrt {4 + {e^2}} }}\)
B.	\(\frac{{\left( {2e - 1} \right)}}{{2\sqrt {4 + {e^2}} }}\)
C.	\(\frac{{\left( {1 + 2e} \right)}}{{\sqrt {4 + {e^2}} }}\)
D.	\(\frac{e}{{\sqrt {4 + {e^2}} }}\)
Answer» C. \(\frac{{\left( {1 + 2e} \right)}}{{\sqrt {4 + {e^2}} }}\)

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If x loge (loge x) – x2 + y2 = 4 (y > 0), then dy/dx at x = e is equal to:

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