Derivative No. 2

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Each $x$ in the product rule has been replaced by $r$.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

Related Posts

Christina Daniel

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.

Leave a Reply