JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

If $x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4 (y > 0)$ then dy/dx at x = e is equal to :

A
$\frac{( 2 e - 1 )}{2 4 + e ^{2}}$
B
$\frac{( 1 + 2 e )}{2 4 + e ^{2}}$
C
$\frac{( 1 + 2 e )}{4 + e ^{2}}$
D
$\frac{e}{4 + e ^{2}}$

Solution

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