Answer
The Derivative is:
$r'=\frac{\theta-1}{\theta^{\frac{3}{2}}}$
Work Step by Step
$r=2(\frac{1}{\sqrt{\theta}}+\sqrt{\theta})$
$r=2(\frac{1\space+\space\theta}{\sqrt{\theta}})$
Using Quotient Rule to find the Derivative:
$y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$
$r'=2(\frac{(0+(1)\theta^{1-1})(\sqrt{\theta})-(1+\theta)((\frac{1}{2})\theta^{\frac{1}{2}-1})}{(\sqrt{\theta})^2})$
$r'=2(\frac{\sqrt{\theta}-(1+\theta)(\frac{1}{2\sqrt{\theta}})}{\theta})$
$r'=2(\frac{\sqrt{\theta}-\frac{1-\theta}{2\sqrt{\theta}})}{\theta}$
$r'=2(\frac{\frac{2\theta-1-\theta}{2\sqrt{\theta}}}{\theta})$
$r'=2(\frac{\theta-1}{2\theta\sqrt{\theta}})$
$r'=\frac{\theta-1}{\theta\sqrt{\theta}}$
$r'=\frac{\theta-1}{\theta^{\frac{3}{2}}}$
