Question

For $x>o,$ if $(2x{)}^{2y}=4e^{2x-2y},$ then $(1+lo{g}_{e}2x{)}^2\frac{dy}{dx}$ is equal to

• A. $lo{g}_{e}2x$
• B. $\frac{xlo{g}_{e}2x-lo{g}_{e}2}{x}$
• C. $xlo{g}_{e}2x$
• D. $\frac{xlo{g}_{e}2x+lo{g}_{e}2}{x}$

Answer 1

The correct option is D $\frac{xlo{g}_{e}2x+lo{g}_{e}2}{x}$

For $x>0(2x{)}^{2y}=4e^{2x-2y}$

taking the log with the side we get

$2ylo{g}_{e}2x=2lo{g}_{e}2+(2x-2y)lo{g}_{e}e$

$2ylo{g}_{e}2x=2lo{g}_{e}2+2x-2y$

$ylo{g}_{e}2x=lo{g}_{e}2+x-y$ ………... $\left(1\right)$

on differentiating equation $\left(1\right)$ w.r.t x we get,

$lo{g}_{e}2x\frac{dy}{dx}+\frac{y}{x}=0+1-\frac{dy}{dx}$

$\frac{dy}{dx}[1+lo{g}_{e}2x]=1-\frac{y}{x}$ ………. $\left(2\right)$

from equation $\left(1\right)$

$y[1+lo{g}_{e}2x]=x+lo{g}_{e}2$

$y=\frac{x+lo{g}_{e}2}{1+lo{g}_{e}2x}$

from equation $\left(2\right)$

$(1+lo{g}_{e}2x)\frac{dy}{dx}=1-\frac{x+lo{g}_{e}2}{x(1+lo{g}_{e}2x)}$

$(1+lo{g}_{e}2x{)}^2\frac{dy}{dx}=\frac{x+xlo{g}_{e}2x-x-lo{g}_{e}2}{x}$

$(1+lo{g}_{e}2x{)}^2\frac{dy}{dx}=\frac{xlo{g}_{e}2x-lo{g}_{e}2}{x}$.