Christina DanielUsing the methods in this post, I would like to evaluate
$\frac{dz(r)}{dr}\bigg|_{r^+}$
with
$z(r)=r\cos\theta$.
Substituting, the expression to evaluate is
$ \frac{d (r \cos\theta)}{dr} \bigg|_{r^+}$.
From the product rule,
$ \frac{dz(r)}{dr} \bigg|_{r^+} = r \frac{d( \cos\theta )}{dr}\big|_{r^+} + \cos\theta \frac{dr}{dr}\big|_{r^+}$
Each $x$ in the product rule has been replaced by $r$.
Since $ \cos\theta $ does not depend on $r$, $ \cos\theta $ is a constant function with respect to $r$. From this post, it follows that
$\frac{d ( \cos\theta ) }{dr} \bigg|_{r^+} = 0$, so
$ \frac{dz(r)}{dr} \bigg|_{r^+} = \cos\theta \frac{dr}{dr}\big|_{r^+} $.
In this post, it was shown that $ \frac{dr}{dr} \big|_{r^+} = 1 $, so
$\boxed{ \frac{dz(r)}{dr} \bigg|_{r^+} = \cos\theta }$.
That’s it!