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If $x=e^{y+e^{y+. .10 \infty}}, x>0$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{1-x}{x}$
$\frac{1-x}{x}$
$x=e^{y+e^{y+e^{y+\ldots \ldots \infty}}} \Rightarrow x=e^{y+x}$
$\Rightarrow \ln x-x=y \Rightarrow \frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x}$
$\Rightarrow \ln x-x=y \Rightarrow \frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x}$
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