Christina DanielUsing the methods in this post, I would like to evaluate
$\frac{dy(\theta)}{d\theta}\bigg|_{\theta^+}$
with
$y(\theta)=r\sin\theta\sin\phi$
Substituting, the expression to evaluate is
$ \frac{d \sin \theta r \sin \phi }{d\theta} \bigg|_{\theta^+}$.
From the product rule,
$ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = \sin\theta \frac{d r \sin \phi }{d\theta}\big|_{\theta^+} + r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+}$
Since $r \sin \phi $ does not depend on $\theta$, $r \sin \phi $ is a constant function with respect to $\theta$. From this post, it follows that
$\frac{dr \sin \phi }{d\theta} \bigg|_{\theta^+} = 0$, so
$ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+} $
In this post, it was shown that
$ \frac{d \sin\theta}{d\theta} \bigg|_{\theta^+} = \cos\theta $.
In the current post, the independent variable is $\theta$ instead of $x$.
Therefore,
$ \boxed{ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = r \cos\theta \sin \phi = z(r,\theta,\phi) \sin \phi } $.