Question

If $y={\log }_e(x+{\log }_e(x+....\left)\right),$ then $\frac{dy}{dx}$ at $(x=e^2-2,y=\sqrt{2})$ is

• A. $\frac{1}{e^{\sqrt{2}}-1}$
• B. $\frac{\log 2}{2\sqrt{2}(e^2-1)}$
• C. $\frac{\sqrt{2}\log \frac{e}{2}}{(e^2-1)}$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{e^{\sqrt{2}}-1}$

Given,

${\log }_e(x+{\log }_e^{(x+\dots )})$

$\therefore y={\log }_e^{x+y}$ $\because$ (as goes on to infinity)

$e^y=x+y$

Differentiate on both sides

$e^y\frac{dy}{dx}=1+\frac{dy}{dx}$

$\because \frac{dy}{dx}=\frac{1}{e^y-1}$

$\because at(x=e^2-2,y=\sqrt{2})=\frac{1}{e^{\sqrt{2}}-1}$