If (2x)2y=4(e2x−2y) then value of (dxdy)(loge2x+1)2 is equal to (A) xxln2x+ln2 (B) xxln2x−ln2 (C) 2xxln2x−ln2 (D) 2xxln2x+ln2

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

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

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

then value of

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

is equal to (A)

\frac{x ln 2 x + ln 2}{x}

(B)

\frac{x ln 2 x - ln 2}{x}

(C)

\frac{x ln 2 x - ln 2}{2 x}

(D)

\frac{x ln 2 x + ln 2}{2 x}

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Question Text	If $(2 x)^{2 y} = 4 (e^{2 x - 2 y})$ then value of $(\frac{d y}{d x}) (lo g_{e} 2 x + 1)^{2}$ is equal to (A) $\frac{x ln 2 x + ln 2}{x}$ (B) $\frac{x ln 2 x - ln 2}{x}$ (C) $\frac{x ln 2 x - ln 2}{2 x}$ (D) $\frac{x ln 2 x + ln 2}{2 x}$
Answer Type	Video solution: 1
Upvotes	150