# Derivative No. 4

Using the methods in this post, I would like to evaluate

$\frac{dz}{d\phi}\bigg|_{\phi^+}$

with

$z=r\cos\theta$.

Substituting, the expression to evaluate is

$\frac{d (r \cos\theta )}{d\phi} \bigg|_{\phi^+}$.

Since $r \cos\theta$ does not depend on $\phi$, $r \cos\theta$ is a constant function with respect to $\phi$. From this post, it follows that

$\boxed { \frac{dz}{d\phi} \bigg|_{\phi^+} = 0 }$.

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#### 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.