If \(x=e^{y+e^{y+e^{y+\ldots}}}\), then \(\frac{d y}{d x}=\)
If \(x=e^{y+e^{y+e^{y+\ldots}}}\), then \(\frac{d y}{d x}=\)
\(\frac{1-x}{x}\)
\(\frac{1}{x}\)
\(\frac{x}{1+x}\)
\(\frac{1+x}{x}\)
Solution
It is given that,
\(\begin{array}{lll}
& x =e^{y+e^{y+e^{y+\ldots}}} \\
\Rightarrow & x =e^{y+x} \\
\Rightarrow & \log _e x =x+y \Rightarrow y=\log _e x-x
\end{array}\)
On differentiating both sides with respect to ' \(x\) ', we get
\(\frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x}\)
Hence, option (a) is correct.