Christina DanielUsing the methods in this post, I would like to evaluate
$\frac{dy(r)}{dr}\bigg|_{r^+}$
with
$y(r)=r \sin\theta\sin\phi$.
This method is not the traditional method taught in a Calculus course because I only want to use properties and rules that I have derived myself on this website.
Substituting, the expression to evaluate is
$ \frac{d ( r \sin \theta \sin \phi )}{dr} \bigg|_{r^+}$.
From the product rule,
$ \frac{dy(r)}{dr} \bigg|_{r^+} = r \frac{d(\sin \theta \sin \phi)}{dr}\big|_{r^+} + \sin \theta \sin \phi \frac{dr}{dr}\big|_{r^+}$
Each $x$ in the product rule has been replaced by $r$.
Since $ \sin \theta \sin \phi $ does not depend on $r$, $ \sin \theta \sin \phi $ is a constant function with respect to $r$. From this post, it follows that
$\frac{d (\sin \theta \sin \phi ) }{dr} \bigg|_{r^+} = 0$, so
$ \frac{dy(r)}{dr} \bigg|_{r^+} = \sin \theta \sin \phi \frac{dr}{dr}\big|_{r^+} $.
In this post, it was shown that $ \frac{dr}{dr} \big|_{r^+} = 1 $, so
$\boxed{ \frac{dy(r)}{dr} \bigg|_{r^+} = \sin \theta \sin \phi }$.
That wasn’t so bad, right?