Derivative No. 3

Using 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Each $x$ in the product rule has been replaced by $r$.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!

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I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.

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