If y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\), then \(\rm \dfrac{dy}{dx}\) is:

Concept:

Chain Rule of Derivatives: 

\(\rm \dfrac{d}{dx}f(g(x))=\dfrac{d}{d\ g(x)}f(g(x))\times \dfrac{d}{dx}g(x)\).

\(\rm \dfrac{d}{dx}e^x\) = e^x.

Calculation:

It is given that y = \(\rm e^{x+e^{x+e^{x+^{\ ...\ \infty}}}}\).

∴ y = \(\rm e^{x+(e^{x+e^{x+^{\ ...\ \infty}}})}=e^{x+y}\)

Differentiating both sides with respect to x and using the chain rule, we get:

\(\rm \dfrac{dy}{dx}=\dfrac{d}{dx}e^{x+y}\)

⇒ \(\rm \dfrac{dy}{dx}=e^{x+y}\dfrac{d}{dx}(x+y)\)

⇒ \(\rm \dfrac{dy}{dx}=y\left (1+\dfrac{dy}{dx} \right )\)

⇒ \(\rm \dfrac{dy}{dx}=y+y\dfrac{dy}{dx}\)

⇒ \(\rm (1-y)\dfrac{dy}{dx}=y\)

⇒ \(\rm \dfrac{dy}{dx}=\dfrac{y}{1-y}\).