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If $x=e^{(y+e)^{(y+e)}(y+\ldots \ldots \infty)}$, then $\frac{d y}{d x}=$
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2868 Upvotes
Verified Answer
The correct answer is:
$\frac{1-x}{x}$
Here $x=e^{y+x}$
Differentiating w.r.t. $\mathrm{x}$
$$
\begin{array}{l}
1=e^{y+x}\left(\frac{d y}{d x}+1\right) \Rightarrow 1=e^{y+x} \frac{d y}{d x}+e^{y+x} \\
\therefore 1=x \cdot \frac{d y}{d x}+x \\
\frac{d y}{d x}=\frac{1-x}{x}
\end{array}
$$
Differentiating w.r.t. $\mathrm{x}$
$$
\begin{array}{l}
1=e^{y+x}\left(\frac{d y}{d x}+1\right) \Rightarrow 1=e^{y+x} \frac{d y}{d x}+e^{y+x} \\
\therefore 1=x \cdot \frac{d y}{d x}+x \\
\frac{d y}{d x}=\frac{1-x}{x}
\end{array}
$$
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