For x - 1 - if 2 x 2 y - 4 e 2 x - 2 y - then 1 - log e 2 x 2 dy dx is equal to -

The correct option is D x log _ e 2 x log _ e 2 x ( 2 x ) ^ 2 y = 4 e ^ 2 x 2 y 2 y #160; ln 2 x = ln 4 + 2 x 2 y y = x ...

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Chain Rule of Differentiation

For x > 1 ,...

Question

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

{log}_{e} 2 x

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\frac{x {log}_{e} 2 x + {log}_{e} 2}{x}

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x {log}_{e} 2 x

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\frac{x {log}_{e} 2 x - {log}_{e} 2}{x}

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{| x - 1 |}^{{log}_{e}^{2} x - {log}_{e} x^{2}} = {| x - 1 |}^{3}

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x

x > 1

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

(1 + {log}_{e} 2 x)^{2} \frac{d y}{d x}

f (x) = {cos}^{- 1} [\frac{1 - (l o g x)^{2}}{1 + (l o g x)^{2}}]

f^{'} (e)

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