Question

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

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

Answer 1

The correct option is C $\frac{1-x}{x}$

Given, $x=e^{y+e^{y+...}}$

$\Rightarrow x=e^{y+x}$

Taking $\log$ both sides

$\log x=(y+x)$

Differentiating both sides w.r.t $x$, we get

$\Rightarrow \frac{1}{x}=\frac{dy}{dx}+1$

$\Rightarrow \frac{dy}{dx}=\frac{1-x}{x}$