# Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$\mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right)$ and $\mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

Solution:

$\operatorname{as} x=e^{\theta}\left(\theta+\frac{1}{\theta}\right)$

Differentiating it with respect to $\theta$ using the product rule,

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{e}^{\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{\theta}\right)$

$=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\frac{\theta^{2}+1}{\theta}\left(\mathrm{e}^{\theta}\right)$

$=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}+\frac{\theta^{2}+1}{\theta}\right)$

$=\mathrm{e}^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)$

$\frac{d x}{d \theta}=e^{\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right) \ldots \ldots(1)$

And, $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

Differentiating it with respect to $\theta$ using the product rule,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{e}^{-\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{-\theta}\right)$

$=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)+\left(\theta-\frac{1}{\theta}\right) \mathrm{e}^{-\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}(-\theta)$

$=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)+\left(\theta-\frac{1}{\theta}\right) \mathrm{e}^{-\theta}(-1)$

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right)$

$=\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)$

$\frac{d y}{d \theta}=e^{-\theta}\left(\frac{-\theta^{2}+\theta^{2}+\theta+1}{\theta^{2}}\right) \ldots \ldots(2)$

divide equation (2)by (1)

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\mathrm{e}^{-\theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right) \times \frac{1}{\mathrm{e}^{\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right)}$

$=e^{-2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)$