# Derivative No. 5

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

$\frac{dy(\phi)}{d\phi}\bigg|_{\phi^+}$

with

$y(\phi)=r\sin\theta\sin\phi$.

Substituting, the expression to evaluate is

$\frac{d r \sin \theta \sin \phi }{d\phi} \bigg|_{\phi^+}$.

From the product rule,

$\frac{dy(\phi)}{d\phi} \bigg|_{\phi^+} = r \sin \theta \frac{d \sin \phi }{d\phi}\big|_{\phi^+} + \sin \phi \frac{d r \sin \theta }{d\phi}\big|_{\phi^+}$

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

$\frac{d r \sin \theta }{d\phi} \bigg|_{\phi^+} = 0$, so

$\frac{dy(\phi)}{d\phi} \bigg|_{\phi^+} = r \sin \theta \frac{d \sin \phi }{d\phi}\big|_{\phi^+}$

In this post, it was shown that

$\frac{d \sin\phi}{d\phi} \bigg|_{\phi^+} = \cos\phi$.

In the current post, the independent variable is $\phi$ instead of $x$.

Therefore,

$\boxed{ \frac{dy(\phi)}{d\phi} \bigg|_{\phi^+} = r \sin \theta \cos\phi = x(r,\theta,\phi) }$.

It is interesting that the derivative of y is x, in this case.

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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.