If $x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0)$, then $\frac{d y}{d x}$ at $x=e$ is equal to:

(B) $\frac{(2 \mathrm{e}-1)}{2 \sqrt{4+\mathrm{e}^{2}}}$

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

(C) $\frac{(1+2 \mathrm{e})}{\sqrt{4+\mathrm{e}^{2}}}$

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

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Differentiation

x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4 (y > 0)

, then

\frac{d y}{d x}

x = e

is equal to:

Answer: (B)

Differentiate

x ln (ln x) - x^{2} + y^{2} = 4

implicitly:

\frac{d}{d x} (x ln (ln x)) = ln (ln x) + \frac{1}{ln x}

, so

ln (ln x) + \frac{1}{ln x} - 2 x + 2 y \frac{d y}{d x} = 0

. At

x = e

ln (ln e) = ln 1 = 0

and

\frac{1}{ln e} = 1

, leaving

1 - 2 e + 2 y \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = \frac{2 e - 1}{2 y}

. The original equation at

x = e

gives

y^{2} = 4 + e^{2}

, so

y = 4 + e^{2}

and

\frac{d y}{d x} = \frac{2 e - 1}{2 4 + e ^{2}}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option