Question

If $x=e^{y+e^{y+......to\infty }},x>0,$ the $\frac{dy}{dx}$ is

• A. $\frac{x}{1+x}$
• B. $\frac{1}{x}$
• C. $\frac{1-x}{x}$
• D. $\frac{1+x}{x}$

Answer 1

Answer: (C)

$x=e^{y+e^{y+.....\infty }}$ $\therefore$ $x=e^{y+x}$ Taking log on both sides, we get $\log x=(y+x)$ Differentiate w.r. to $x,$ we get $\frac{1}{x}=\frac{dy}{dx}+1\Rightarrow \frac{dy}{dx}=\frac{1-x}{x}$