Answer
$$\frac{{dr}}{{d\theta }} = \frac{1}{{\root 3 \of \theta }} + \frac{1}{{\root 4 \of \theta }} + \frac{1}{{\sqrt \theta }}$$
Work Step by Step
$$\eqalign{
& r - 2\sqrt \theta = \frac{3}{2}{\theta ^{2/3}} + \frac{4}{3}{\theta ^{3/4}} \cr
& {\text{differentiate both sides with respect to }}\theta \cr
& \frac{d}{{d\theta }}\left( r \right) - \frac{d}{{d\theta }}\left( {2\sqrt \theta } \right) = \frac{d}{{d\theta }}\left( {\frac{3}{2}{\theta ^{2/3}}} \right) + \frac{d}{{d\theta }}\left( {\frac{4}{3}{\theta ^{3/4}}} \right) \cr
& \cr
& {\text{Find derivatives}} \cr
& \frac{{dr}}{{d\theta }} - 2\left( {\frac{1}{{2\sqrt \theta }}} \right) = \frac{3}{2}\left( {\frac{2}{3}{\theta ^{ - 1/3}}} \right) + \frac{4}{3}\frac{d}{{d\theta }}\left( {\frac{3}{4}{\theta ^{ - 1/4}}} \right) \cr
& \frac{{dr}}{{d\theta }} - \frac{1}{{\sqrt \theta }} = {\theta ^{ - 1/3}} + {\theta ^{ - 1/4}} \cr
& \cr
& {\text{Solve for }}\frac{{dr}}{{d\theta }} \cr
& \frac{{dr}}{{d\theta }} = {\theta ^{ - 1/3}} + {\theta ^{ - 1/4}} + \frac{1}{{\sqrt \theta }} \cr
& \frac{{dr}}{{d\theta }} = \frac{1}{{\root 3 \of \theta }} + \frac{1}{{\root 4 \of \theta }} + \frac{1}{{\sqrt \theta }} \cr} $$
