If $x=e^{y+e^{y+ ........ to \,\, ∞}}, x> 0$ then $\frac{d^2y}{dx^2}$ is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-\frac{1}{x^2}$
$x=e^{y+e^{y+ ........ to \,\, ∞}}$
$x=e^{y+x}⇒\log x=y+x$
so $y=\log x=x$
$\frac{dy}{dx}=\frac{1}{x}-1⇒\frac{d^2y}{dx^2}=-\frac{1}{x^2}$