If xloge(logex)−x2+y2=4 then (dxdy)atx=e is equal to (A) 4+e22e+1 (B) 24+e2e (C) 2(4+e2)2e+1 (D) 2(4+e2)2e−1

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If {x}{{}_{{e}}{({{}_{{e}}{x}})}}-{x}^{{2}}+{y}^{{2}}={4} then

Question

If $x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4$ then $(\frac{d y}{d x})_{a t x = e}$ is equal to (A) $\frac{2 e + 1}{4 + e ^{2}}$ (B) $\frac{e}{2 4 + e ^{2}}$ (C) $\frac{2 e + 1}{2 ( 4 + e ^{2} )}$ (D) $\frac{2 e - 1}{2 ( 4 + e ^{2} )}$

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Scarlett Discussed

x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4

then

(\frac{d y}{d x})_{a t x = e}

is equal to (A)

\frac{2 e + 1}{4 + e ^{2}}

(B)

\frac{e}{2 4 + e ^{2}}

(C)

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

(D)

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

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Question Text

x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4

then

(\frac{d y}{d x})_{a t x = e}

is equal to (A)

\frac{2 e + 1}{4 + e ^{2}}

(B)

\frac{e}{2 4 + e ^{2}}

(C)

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

(D)

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