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