If y=ex+ex+ex⋯∞, then dydx=

y=ex+y⇒log⁡y=(x+y)log⁡e=x+y⇒1y⋅dydx=1+dydx⇒dydx1y−1=1⇒dydx1−yy=1⇒dydx=y1−y

If y=ex+ex+ex⋯∞, then dydx=

y=ex+y⇒log⁡y=(x+y)log⁡e=x+y⇒1y⋅dydx=1+dydx⇒dydx1y−1=1⇒dydx1−yy=1⇒dydx=y1−y

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If $y = e^{x + e^{x + e^{x \dots \infty}}}$ , then $\frac{dy}{dx} =$

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$y$

$\frac{1}{y}$

$\frac{1}{1 - y}$

$\frac{y}{1 - y}$

answer is D.

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Detailed Solution

$\begin{array}{l} y = e^{x + y} \Rightarrow \log y = (x + y) \log e = x + y \\ \Rightarrow \frac{1}{y} \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx} \Rightarrow \frac{dy}{dx} (\frac{1}{y} - 1) = 1 \Rightarrow \frac{dy}{dx} (\frac{1 - y}{y}) = 1 \Rightarrow \frac{dy}{dx} = \frac{y}{1 - y} \end{array}$