Question

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

Answer 1

$\because x=e^{\theta }(\theta +\frac{1}{\theta })andy=e^{\theta }(\theta +\frac{1}{\theta })\therefore \frac{dx}{d\theta }=\frac{d}{d\theta }\left[e^{\theta }(\theta +\frac{1}{\theta })\right]=e^{\theta }.\frac{d}{d\theta }(\theta +\frac{1}{\theta })+(\theta +\frac{1}{\theta })\frac{d}{d\theta }{e}^{\theta }=e^{\theta }(1-\frac{1}{{\theta }^2})+(\theta +\frac{1}{{\theta }^2})e^{\theta }=e^{\theta }\left(\frac{{\theta }^2-1+{\theta }^3+\theta }{{\theta }^2}\right)\dots \left(i\right)$

and $\frac{dy}{d\theta }=\frac{d}{d\theta }\left[e^{\theta }(\theta -\frac{1}{\theta })\right]=e^{\theta }.\frac{d}{d\theta }(\theta -\frac{1}{\theta })+\frac{d}{d\theta }{e}^{\theta }(\theta -\frac{1}{\theta })=e^{\theta }(1-\frac{1}{{\theta }^2})+(\theta +\frac{1}{\theta })e^{\theta }\frac{d}{d\theta }(-\theta )=e^{\theta }[\frac{{\theta }^2+1}{{\theta }^2}-\frac{{\theta }^{2}11}{\theta }]=e^{|}-\theta \left[\frac{{\theta }^2+1-{\theta }^3+\theta }{{\theta }^2}\right]\dots \left(ii\right)\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{e^{\theta }\left(\frac{{\theta }^2+1-{\theta }^3+\theta }{{\theta }^2}\right)}{e^{\theta }\left(\frac{{\theta }^2-1+{\theta }^3+\theta }{{\theta }^2}\right)}=e^{-2\theta }\left(\frac{-{\theta }^3+{\theta }^2+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right)$