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Each $x$ in the product rule has been replaced by $r$.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