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According to my book the following function has a vanishing partial derivative with respect to $\phi$:

$$f(\phi(\theta), \phi^{'}(\theta), \theta)\ = \sqrt {1+sin^2(\theta)(\phi^{'})^2} $$

I can see that there is no explicit $\phi$-dependence, but I feel like I should be able to form the following expression using the chain rule:

$$\frac{\partial f}{\partial \phi}=\frac{\partial f}{\partial \phi^{'}}\frac{d \phi^{'}}{d \theta}\frac{d \theta}{d \phi}=\frac{\partial f}{\partial \phi^{'}}\frac{d^{2} \phi}{d \theta^{2}}\frac{1}{\phi^{'}}$$

It seems I have a fundamental misconception here, can anyone help me see what I'm missing?

FieldOfDreams
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    To take the derivative with respect to $\phi$ means to holds all other variables constant, so you can't compute $d\phi'/d\theta$ (as an example) as part of the calculation. – march Oct 14 '21 at 17:43
  • @march That helps, thanks. I also tried relabeling the variables and it became quite easy to see. It seems my confusion arises from the interdependence of $\phi^{'}$, $\theta$, and $\phi$. It seems like varying $\phi$ should feedback on the other two variables and in doing so, change the value of $f$. – FieldOfDreams Oct 14 '21 at 18:07
  • Related: https://math.stackexchange.com/questions/1963640/why-does-fracdqdt-not-depend-on-q-why-does-the-calculus-of-variations, https://math.stackexchange.com/questions/580858/why-is-frac-operatorname-dy-operatorname-dy-zero-since-y-depends-on, and many other questions (see the lists of linked questions there). – Hans Lundmark Oct 14 '21 at 20:09

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