We know that azimuth ($\theta$) and elevation ($\phi$) angles can represent a unit vector as $\mathbf{e}=\begin{bmatrix}\cos\theta\cos\phi \\ \sin\theta\cos\phi \\ \sin\phi\end{bmatrix}$. It is easy to calculate the jacobian of unit vector with respect to the angles, but the opposite seems difficult.
How to determine $\mathbf{J}_\theta=\frac{\partial\theta}{\partial\mathbf{e}}$ and $\mathbf{J}_\phi=\frac{\partial\phi}{\partial\mathbf{e}}$? These should be matrices with the size of $1\times3$, but I am not sure how to calculate them. One way could be that we invert the relation of coordinates and calculate the derivative by each coordinate of $\mathbf{d}=\begin{bmatrix}d_1 & d_2 & d_3]\end{bmatrix}^T$. For example for the 1st coordinate it yields $\theta =\arccos(\frac{d_1}{\cos\phi}) \Rightarrow \mathbf{J}_{\theta_{11}}=\frac{\mathrm{d}\arccos(\frac{d_1}{\cos\phi})}{\mathrm{d}d_1}$, but I am not sure this is the correct way.
The other way what I see inverting the full relation (not just by coordinates) and somehow find $\theta=f(\mathbf{e})$ relation, then the derivative of the expression should be determined by each coordinate. My problem with this method is that how to calculate $f(.)$? I cant give a linear expression for f(.), because the closed formula requires the division of the coordinates of $\mathbf{d}$ and the usage of trigonometric functions.
Can someone clear me and tell me the correct way to do this?