If x = e^θ ≤ft( θ + 1/(θ) ) and y = e^-θ ≤ft( θ - 1/(θ) ), then dy/dx…

The correct answer is Option (4) → $e^{-2\theta} \left( \frac{1 + \theta + \theta^2 - \theta^3}{\theta^3 + \theta^2 + \theta - 1} \right)$ ##

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

Taking derivative of both equations w.r.t. $\theta$, we get

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

$= e^\theta \cdot \frac{d}{d\theta} \left( \theta + \frac{1}{\theta} \right) + \left( \theta + \frac{1}{\theta} \right) \cdot \frac{d}{d\theta} e^\theta \quad \text{[by product rule]}$

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

$= e^\theta \left[ \frac{\theta^2 - 1 + \theta^3 + \theta}{\theta^2} \right] \quad \dots(i)$

and

$\frac{dy}{d\theta} = \frac{d}{d\theta} \left[ e^{-\theta} \cdot \left( \theta - \frac{1}{\theta} \right) \right]$

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

$= e^{-\theta} \left( 1 + \frac{1}{\theta^2} \right) + \left( \theta - \frac{1}{\theta} \right) e^{-\theta} \cdot \frac{d}{d\theta} (-\theta) \quad \text{[by product rule]}$

$= e^{-\theta} \left[ \frac{\theta^2 + 1}{\theta^2} - \frac{\theta^2 - 1}{\theta} \right] = e^{-\theta} \left[ \frac{\theta^2 + 1 - \theta^3 + \theta}{\theta^2} \right] \quad \dots(ii)$

On dividing Eq. (ii) by Eq. (i), we get

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