If \( y=e^{x+e^{x+e^{x+\cdots\infty}}} \) then \( \displaystyle\frac{dy}{dx}= \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{y}{1-y} \)
The exponent is \( x \) plus the whole expression again, so the infinite tower gives
\( y=e^{x+y} \)
Differentiating with respect to \( x \):
\( \displaystyle \frac{dy}{dx}=e^{x+y}\left(1+\frac{dy}{dx}\right) \)
\( \displaystyle =y\left(1+\frac{dy}{dx}\right) \)
Collecting the terms in \( \frac{dy}{dx} \):
\( \displaystyle \frac{dy}{dx}(1-y)=y \)
\( \displaystyle \frac{dy}{dx}=\frac{y}{1-y} \)