Answer
$h_p=\sinΦ \cos \theta$;
$h_Φ=p \cosΦ \cos \theta$;
$h_{\theta}=-p \sinΦ \sin \theta$
Work Step by Step
Our aim is to take the first partial derivative of the given function $h$ with respect to $p$, treating $Φ$ and $θ$ as constants:
$h_p=\sinΦ \cos \theta$
Now, we repeat the process for the other partial derivatives:
$h_Φ=p \cosΦ \cos \theta$
$h_{\theta}=-p \sinΦ \sin \theta$
