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If, $x=e^\theta\left(\theta+\frac{1}{\theta}\right), y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$, then find $\frac{d y}{d x}$.
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$\because x=e^\theta\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$
$$
\begin{aligned}
\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=& \frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^\theta \cdot\left(\theta+\frac{1}{\theta}\right)\right] \\
&=\mathrm{e}^\theta \cdot \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}^\theta} \mathrm{e}^\theta \\
&=\mathrm{e}^\theta\left(1-\frac{1}{\theta^2}\right)+\left(\theta+\frac{1}{\theta}\right) \mathrm{e}^\theta \\
&=\mathrm{e}^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)
\end{aligned}
$$
and $\begin{aligned} \frac{d y}{d \theta} &=\frac{d}{d \theta}\left[e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right] \\ &=e^{-\theta} \cdot \frac{d}{d \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{d}{d \theta} e^{-\theta} \\ &=e^{-\theta}\left(1+\frac{1}{\theta^2}\right)+\left(\theta-\frac{1}{\theta}\right) \mathrm{e}^{-\theta} \cdot \frac{d}{d \theta}(-\theta) \\ &=e^{-\theta}\left[\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right] \\ \therefore \quad \frac{d y}{d x} &=\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}\left(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right)}{e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)} \end{aligned}$
$=\mathrm{e}^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right)$
$$
\begin{aligned}
\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=& \frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^\theta \cdot\left(\theta+\frac{1}{\theta}\right)\right] \\
&=\mathrm{e}^\theta \cdot \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}^\theta} \mathrm{e}^\theta \\
&=\mathrm{e}^\theta\left(1-\frac{1}{\theta^2}\right)+\left(\theta+\frac{1}{\theta}\right) \mathrm{e}^\theta \\
&=\mathrm{e}^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)
\end{aligned}
$$
and $\begin{aligned} \frac{d y}{d \theta} &=\frac{d}{d \theta}\left[e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right] \\ &=e^{-\theta} \cdot \frac{d}{d \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{d}{d \theta} e^{-\theta} \\ &=e^{-\theta}\left(1+\frac{1}{\theta^2}\right)+\left(\theta-\frac{1}{\theta}\right) \mathrm{e}^{-\theta} \cdot \frac{d}{d \theta}(-\theta) \\ &=e^{-\theta}\left[\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right] \\ \therefore \quad \frac{d y}{d x} &=\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}\left(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right)}{e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)} \end{aligned}$
$=\mathrm{e}^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right)$
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