Question

If $x=e^{y+e^y+\cdots \infty },x>0$, 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

Answer: (C)

$\frac{1-x}{x}$

Given, $x=e^{y+e^{y+e^y+\cdots \infty }}$
$\therefore x=e^{y+x}$
Taking $\log$ on both sides, we get
$\log x=(y+x)\log e$
$\Rightarrow \log x=y+x$
On differentiating with respect to $x$, we get
$\frac{1}{x}=\frac{dy}{dx}+1$
$\Rightarrow \frac{dy}{dx}=\frac{1}{x}-1=\frac{1-x}{x}$