Evaluation of ${\partial k_{x_h}'}/{\partial \phi}$and ${\partial k_{y_h}'}/{\partial \phi}$

Differentiating equation (14) and after some algebraic manipulation, we obtain the following:

$$\begin{eqnarray} \frac{\partial k_{x_h}'}{\partial \phi} = &=& -k_z\tan\gamma \f... ...\tan\gamma \sin \alpha'_y \frac{\partial k_{y_m}'}{\partial \phi}.\end{eqnarray}$$ (15)

Differentiating equation (13) we can write the following:  

$$\frac{\partial k_{y_h}'}{\partial \phi} = k_z\tan\gamma \le... ...{\partial \sin \alpha'_y}{\partial \phi} \tan\alpha'_x \right].$$ (16)

To evaluate equation (16) we need $\partial \tan\alpha'_x/\partial \phi$and $\partial \sin \alpha'_y/\partial \phi$;that is,  

$$\frac{\partial \tan \alpha'_x}{\partial \phi} = \frac{\parti... ... \phi} = \frac{1}{k_z} \frac{\partial k_{x_m}'}{\partial \phi},$$ (17)

and  

$$\frac{\partial \sin \alpha'_y}{\partial \phi} = \frac{\parti... ...}{k_z}\cos^3 \alpha'_y \frac{\partial k_{y_m}'}{\partial \phi}.$$ (18)

Substituting equation (17) and equation (18) into equation (16), we finally obtain:

$$\begin{eqnarray} \frac{\partial k_{y_h}'}{\partial \phi} &=& k_z\tan\gamma \left... ...'_y \frac{\partial k_{y_m}'}{\partial \phi} \right]. \nonumber \\ \end{eqnarray}$$