Derivative No. 7

Using the methods in this post, I would like to evaluate $\frac{dy(\theta)}{d\theta}\bigg|_{\theta^+}$ with $y(\theta)=r\sin\theta\sin\phi$ Substituting, the expression to evaluate is $ \frac{d \sin \theta r \sin \phi  }{d\theta} \bigg|_{\theta^+}$. From the product rule, $ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = \sin\theta \frac{d r \sin \phi  }{d\theta}\big|_{\theta^+} + r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+}$ Since $r \sin \phi $ does not depend on $\theta$, $r \sin \phi $ is a constant function with respect to $\theta$. From this post, it follows that $\frac{dr \sin \phi }{d\theta} \bigg|_{\theta^+} = 0$, so $ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = r \sin \phi \frac{d \sin\theta }{d\theta}\big|_{\theta^+} $ In this post, it was shown that $ \frac{d \sin\theta}{d\theta} \bigg|_{\theta^+} = \cos\theta $. In the current post, the independent variable is $\theta$ instead of $x$. Therefore, $ \boxed{ \frac{dy(\theta)}{d\theta} \bigg|_{\theta^+} = r \cos\theta \sin \phi    = z(r,\theta,\phi) \sin \phi  } $.