Answer
$(sec\sqrt{\theta})[\frac{tan \sqrt{\theta}tan(\frac{1}{\theta})}{2\sqrt{\theta}}-\frac{sec^2(\frac{1}{\theta})}{\theta^2}]$
Work Step by Step
$r=(sec\sqrt{\theta}tan(\frac{1}{\theta}))$
Apply the chain rule:
$\frac{dr}{d\theta}=(sec \sqrt{\theta})(sec^2 \frac{1}{\theta})(\frac{-1}{\theta^2})+tan(\frac{1}{\theta}(sec\sqrt{\theta})tan \sqrt{\theta})(\frac{1}{2\sqrt{\theta}})$
=$\frac{-1}{\theta^2}sec\sqrt{\theta}sec^2(\frac{1}{\theta})+\frac{1}{2\sqrt{\theta}}tan(\frac{1}{\theta})sec \sqrt{\theta} tan \sqrt{\theta}$
=$(sec\sqrt{\theta})[\frac{tan \sqrt{\theta}tan(\frac{1}{\theta})}{2\sqrt{\theta}}-\frac{sec^2(\frac{1}{\theta})}{\theta^2}]$
