Given that \(y = \arctan \sqrt { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in terms of \(x\). Hence show that
$$\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { x } ( x + 1 ) } \mathrm { d } x = \frac { \pi } { 2 }$$
A curve has cartesian equation
$$x ^ { 2 } + y ^ { 2 } = x y + 1$$
Show that the polar equation of the curve is
$$r ^ { 2 } = \frac { 2 } { 2 - \sin 2 \theta }$$
Determine the greatest and least positive values of \(r\) and the values of \(\theta\) between 0 and \(2 \pi\) for which they occur.