Solución
∂
F
∂
θ
=
6
,
∂
F
∂
x
=
4
−
3
3
\frac{\partial F}{\partial \theta}=6,\;\;\;\frac{\partial F}{\partial x}=4−3\sqrt{3}
∂
θ
∂
F
=
6
,
∂
x
∂
F
=
4
−
3
3