For x 1, if (2x)2y = 4e2x - 2y, then ( 1 + loge 2x)2 (dy/dx) is equal to :

Question

For x > 1, if (2x)2y = 4e2x - 2y, then ( 1 + loge 2x)2 (dy/dx) is equal to : (A)  loge 2x (B) (x loge 2x + loge 2/x)  (C) x loge 2x (D) (x loge 2x - loge 2/x)

Tardigrade Admin · Accepted Answer

(2x)2y = 4e2x - 2y 2y ℓ n2x = ℓ n4 + 2x - 2y y = (x + ℓ n 2/1 + ℓ n 2 x) y' = ((1 + ℓ n 2 x) - (x + ℓ n 2 ) (1/x)/( 1+ ℓ n 2x )2) y ' = 1 + ℓ n 2 x)2 = [(x ℓ n 2x - ℓ n 2/x) ]

Q. 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 :

$lo g_{e} 2 x$

$\frac{x l o g _{e} 2 x + l o g _{e} 2}{x}$

$x lo g_{e} 2 x$

$\frac{x l o g _{e} 2 x - l o g _{e} 2}{x}$

$(2 x)^{2 y} = 4 e^{2 x - 2 y}$
$2 y ℓ n 2 x = ℓ n 4 + 2 x - 2 y$
$y = \frac{x + ℓ n 2}{1 + ℓ n 2 x}$
$y^{'} = \frac{( 1 + ℓ n 2 x ) - ( x + ℓ n 2 ) \frac{1}{x}}{( 1 + ℓ n 2 x ) ^{2}}$
$y^{'} = 1 + ℓ n 2 x)^{2} = [\frac{x ℓ n 2 x - ℓ n 2}{x}]$

Q. For x>1, if (2x)2y=4e2x−2y, then (1+loge​2x)2dxdy​ is equal to :

loge​2x

xxloge​2x+loge​2​

xloge​2x

xxloge​2x−loge​2​