This is my original post.

In the derivation, I come across to the point that I need to decompose $\frac{\partial f}{\partial v_x}$ to obtain $\frac{\partial f}{\partial v}$, $\frac{\partial f}{\partial \theta}$,$\frac{\partial f}{\partial \phi}$ terms.

One SE user pointed me where I made mistake in the derivation.

This is his/her suggestion:

$$ \frac{\partial f}{\partial v_x}=\sin(\theta)\cos(\phi)\frac {\partial f}{\partial v}-\frac1v \cos(\theta)\cos(\phi)\frac {\partial f}{\partial \theta}+\frac{\sin(\phi)}{r\sin(\theta)}\frac{\partial f}{\partial \phi}$$

My question/purpose in this post is to understand how to derive/obtain that.

He suggested me to look into his/her other answer for more detail, which is a very good answer to remind me about the derivative of an inverse function.

But still I cannot understand how to do that.

If I write

$$\frac{\partial v_x}{\partial f_o} = \sin\theta\cos\phi \frac{\partial v}{\partial f_o} + v\cos\theta\cos\phi \frac{\partial \theta}{\partial f_o}- v\sin\theta\sin\phi \frac{\partial \phi}{\partial f_o}$$

and try to find the inverse with the determinant = $-v^2 \sin^2\theta \cos\theta \sin\phi \cos^2\phi$, still I do not get a form close to his answer.


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