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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