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If \(x=e^{y+e^{y+e^{y+\ldots}}}\), then \(\frac{d y}{d x}=\)
Options:
Solution:
1781 Upvotes
Verified Answer
The correct answer is:
\(\frac{1-x}{x}\)
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.
\(\begin{array}{lll}
& x =e^{y+e^{y+e^{y+\ldots}}} \\
\Rightarrow & x =e^{y+x} \\
\Rightarrow & \log _e x =x+y \Rightarrow y=\log _e x-x
\end{array}\)
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.
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