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If $e^{-y} \cdot y=x$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{\mathrm{y}}{\mathrm{x}(1-\mathrm{y})}$
$$
\begin{aligned}
& e^{-y} \cdot y=x \\
& \therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \ldots
\end{aligned}
$$
and
$$
\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}
$$
Now $y=\mathrm{xe}^{\mathrm{y}}$
$$
\begin{aligned}
& \therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \\
& \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1}
\end{aligned}
$$
From (1) and (2), we write
$$
\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}
$$
\begin{aligned}
& e^{-y} \cdot y=x \\
& \therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \ldots
\end{aligned}
$$
and
$$
\mathrm{e}^{\mathrm{y}}=\frac{\mathrm{y}}{\mathrm{x}}
$$
Now $y=\mathrm{xe}^{\mathrm{y}}$
$$
\begin{aligned}
& \therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \\
& \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1}
\end{aligned}
$$
From (1) and (2), we write
$$
\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}
$$
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