Question

If $x=e^{\theta }(\theta +\frac{1}{\theta })$ and $y=e^{-\theta }(\theta +\frac{1}{\theta })$, find $\frac{dy}{dx}$

Answer 1

$x=e^{\theta }(\theta +\frac{1}{\theta })$

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

$\frac{dx}{d\theta }=$ $e^{\theta }\frac{({\theta }^3+\theta +{\theta }^2-1)}{{\theta }^2}$.

$y=e^{-\theta }(\theta +\frac{1}{\theta })$

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

$\frac{dy}{d\theta }$ $=\frac{(-{\theta }^3-\theta +{\theta }^2-1)}{{\theta }^2}$

Now, $\frac{dy}{dx}=e^{-2\theta }\frac{({\theta }^3+\theta +{\theta }^2-1)}{(-{\theta }^3-\theta +{\theta }^2-1)}$