Question

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

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

Answer 1

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

$x=e^{y+e^{y+e^{y+...}}}$
$\Rightarrow x=e^{y+x}$
Take logarithm on both sides and differentiating with respect to $x$
$\therefore \frac{1}{x}=1+\frac{dy}{dx}$ $\Rightarrow$ $\frac{dy}{dx}=\frac{1-x}{x}$