Answer
$\cos \theta+ \sin \theta$ and $r(-\sin \theta+\cos \theta)$
Work Step by Step
Given: $g=r \cos \theta+ r \sin \theta$
Now, $\dfrac{\partial g}{\partial \theta}=\dfrac{\partial }{\partial r} [r \cos \theta+ r \sin \theta]=\dfrac{\partial r}{\partial r} [\cos \theta+ \sin \theta]=\cos \theta+ \sin \theta$
and
$\dfrac{\partial g}{\partial \theta}=\dfrac{\partial }{\partial \theta} [r \cos \theta+ r \sin \theta]=r(-\sin \theta+\cos \theta)$
Hence, $\cos \theta+ \sin \theta$ and $r(-\sin \theta+\cos \theta)$
