Question

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

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

Answer 1

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

We have
$x=e^{y+e^{y+\cdots to\infty }}$
It can be written as,
$x=e^{y+x}$
Taking $\log$ on both the sides
$\log x=(y+x)\log e=y+x$
$\therefore y=\log x-x$
By differentiating w.r. to $x$, we get
$\therefore \frac{dy}{dx}=\frac{1}{x}-1=\frac{1-x}{x}$