Question

for $x>1$, If $(2x{)}^{2y}=4e^{2x-2y}$, then
$(1+{\log }_{e}2x{)}^2\frac{dy}{dx}$ is equal to :

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

Answer 1

The correct option is C $\frac{x{\log }_{e}2x-{\log }_{e}2}{x}$

$(2x{)}^{2y}=4e^{2x-2y}$
On taking $\left(\ln \right)$ operator on both sides we get-
$2y\cdot \ln \left(2x\right)=\ln \left(4\right)+(2x-2y)$
$\Rightarrow y=\frac{x+\ln 2}{1+\ln 2x}$
On differentating w.r.t. $x$
$\frac{dy}{dx}=\frac{(1+\ln 2x)\cdot (1+0)-(x+\ln 2)\left(\frac{1}{2x}\right)\cdot 2}{(1+\ln 2x{)}^2}$
$\Rightarrow (1+\ln 2x{)}^2\cdot \frac{dy}{dx}=1+\ln 2x-\frac{x+\ln 2}{x}$
$\Rightarrow (1+\ln 2x{)}^2\cdot \frac{dy}{dx}=\frac{x\ln 2x-\ln 2}{x}$