Find `(dy)/(dx)` , when `x=e^(theta)(theta+1/theta)` and `y=e^(-theta)(theta-1/theta)`

Explanation

To find \(\frac{dy}{dx}\) where \(x = e^{\theta} \left( \theta + \frac{1}{\theta} \right)\) and \(y = e^{-\theta} \left( \theta - \frac{1}{\theta} \right)\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(\theta\)

Given:
\[ x = e^{\theta} \left( \theta + \frac{1}{\theta} \right) \]

Using the product rule, we differentiate \(x\):
\[ \frac{dx}{d\theta} = e^{\theta} \cdot \frac{d}{d\theta}\left(\theta + \frac{1}{\theta}\right) + \left(\theta + \frac{1}{\theta}\right) \cdot \frac{d}{d\theta}(e^{\theta}) \]

Calculating the derivatives:
\[ \frac{d}{d\theta}\left(\theta + \frac{1}{\theta}\right) = 1 - \frac{1}{\theta^2} \]
\[ \frac{d}{d\theta}(e^{\theta}) = e^{\theta} \]

Substituting back:
\[ \frac{dx}{d\theta} = e^{\theta} \left( 1 - \frac{1}{\theta^2} \right) + \left( \theta + \frac{1}{\theta} \right) e^{\theta} \]
\[ = e^{\theta} \left( 1 - \frac{1}{\theta^2} + \theta + \frac{1}{\theta} \right) \]
\[ = e^{\theta} \left( \theta + 1 + \frac{1}{\theta} - \frac{1}{\theta^2} \right) \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)

Given:
\[ y = e^{-\theta} \left( \theta - \frac{1}{\theta} \right) \]

Using the product rule, we differentiate \(y\):
\[ \frac{dy}{d\theta} = e^{-\theta} \cdot \frac{d}{d\theta}\left(\theta - \frac{1}{\theta}\right) + \left(\theta - \frac{1}{\theta}\right) \cdot \frac{d}{d\theta}(e^{-\theta}) \]

Calculating the derivatives:
\[ \frac{d}{d\theta}\left(\theta - \frac{1}{\theta}\right) = 1 + \frac{1}{\theta^2} \]
\[ \frac{d}{d\theta}(e^{-\theta}) = -e^{-\theta} \]

Substituting back:
\[ \frac{dy}{d\theta} = e^{-\theta} \left( 1 + \frac{1}{\theta^2} \right) - \left( \theta - \frac{1}{\theta} \right) e^{-\theta} \]
\[ = e^{-\theta} \left( 1 + \frac{1}{\theta^2} - \theta + \frac{1}{\theta} \right) \]
\[ = e^{-\theta} \left( 1 - \theta + \frac{1}{\theta} + \frac{1}{\theta^2} \right) \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{e^{-\theta} \left( 1 - \theta + \frac{1}{\theta} + \frac{1}{\theta^2} \right)}{e^{\theta} \left( \theta + 1 + \frac{1}{\theta} - \frac{1}{\theta^2} \right)} \]

This simplifies to:
\[ \frac{dy}{dx} = e^{-2\theta} \cdot \frac{1 - \theta + \frac{1}{\theta} + \frac{1}{\theta^2}}{\theta + 1 + \frac{1}{\theta} - \frac{1}{\theta^2}} \]

### Final Answer
\[ \frac{dy}{dx} = e^{-2\theta} \cdot \frac{1 - \theta + \frac{1}{\theta} + \frac{1}{\theta^2}}{\theta + 1 + \frac{1}{\theta} - \frac{1}{\theta^2}} \]