If $x=e^{y+e^{y+e^{y+... \infty}}}$, x > 0, then $\frac{d y}{d x}$ is equal to
Answer & explanation
Correct answer: option 3
$x=e^{y+e^{y+e^{y+... \infty}}}$
⇒ $\log x = y + e^{y+e^{y+e^{y+... \infty}}}$
⇒ $\log x = y + x$
differentiating wrt x
$\frac{1}{x} = \frac{dy}{dx} + 1 ⇒ \frac{dy}{dx} = \frac{1-x}{x}$
Option: 3