Point Corresponding To $\theta = \frac { \pi } { 4 }$

$\text { Find } \frac { d y } { d x } \text { and } \frac { d ^ { 2 } y } { d x ^ { 2 } } \text { if possible, and find the slope and concavity (if possible) at the }$ point corresponding to $\theta = \frac { \pi } { 4 }$ .
$\begin{array} { l } x = 4 \cos \theta \\y = 4 \sin \theta\end{array}$

A) $\frac { d y } { d x } = \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 1 } { 4 \sin ^ { 3 } \theta }$ ; at $\theta = \frac { \pi } { 4 }$ : slope 1 and concave down
B) $\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta }$ ; at $\theta = \frac { \pi } { 4 }$ : slope $- 1$ and concave down
C) $\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta }$ ; at $\theta = \frac { \pi } { 4 }$ : slope $- 1$ and concave up
D) $\frac { d y } { d x } = \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 1 } { 4 \sin ^ { 3 } \theta }$ ; at $\theta = \frac { \pi } { 4 }$ : slope 1 and concave up
E) $\frac { d y } { d x } = - \cot \theta , \frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 1 } { 4 \sin ^ { 3 } \theta }$ ; at $\theta = \frac { \pi } { 4 }$ : slope of $- 1$ and concave down