For $x>1$, if $(2 x)^{2 y}=4 e^{2 x-2 y}$, then $\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x}$ is equal to:

(A) $\frac{x \log _{e} 2 x-\log _{e} 2}{x}$

(C) $\frac{x \log _{e} 2 x+\log _{e} 2}{x}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Differentiation

For

x > 1

, if

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

, then

(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}

is equal to:

Answer: (A)

Take logarithms of

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

2 y ln (2 x) = ln 4 + 2 x - 2 y

. Gathering the

y

terms,

2 y (1 + ln (2 x)) = 2 ln 2 + 2 x

, so

y = \frac{x + l n 2}{1 + l n ( 2 x )}

. Differentiating by the quotient rule, with

\frac{d}{d x} ln (2 x) = \frac{1}{x}

, and simplifying gives

\frac{d y}{d x} = \frac{l n ( 2 x ) - \frac{l n 2}{x}}{( 1 + l n ( 2 x ) ) ^{2}}

. Multiplying through by

(1 + ln (2 x))^{2}

leaves

ln (2 x) - \frac{l n 2}{x} = \frac{x l n ( 2 x ) - l n 2}{x}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option