I want to find the partial derivatives of the expression for $v_3(\boldsymbol{u})$ with respect to $u_1$, $u_2$ and $u_3$ from the expressions below. Here $\Phi$ denotes the cumulative distribution function of the standard normal probability distribution. I think it should be possible to solve it analytically by employing the chain rule. However, I have a feeling that the expressions quickly will become quite ugly. Any suggestions on how this can be done?
$$h(\boldsymbol{u}) = F_{H_s}^{-1}(\boldsymbol{\Phi}(u_1)) = \alpha[-\ln(1-\Phi(u_1))]^{1/\beta}$$
$$t(\boldsymbol{u}) = F_{T_z|H_s}^{-1}(\boldsymbol{\Phi}(u_2)|h(\boldsymbol{u})) = \exp({\mu(h(\boldsymbol{u})) + \sigma(h(\boldsymbol{u})) u_2})$$
$$v_3(\boldsymbol{u}) = F_{V_3|Tz,H_s}^{-1}(\boldsymbol{\Phi}(u_3)|h(\boldsymbol{u}),t(\boldsymbol{u})) = \sqrt{-2m_0(h(\boldsymbol{u}),t(\boldsymbol{u})) \ln\left(-\frac{2\pi}{\widetilde{T}}\sqrt{\frac{m_0(h(\boldsymbol{u}),t(\boldsymbol{u}))}{m_2(h(\boldsymbol{u}),t(\boldsymbol{u}))}}\ln\boldsymbol{\Phi}(u_3)\right)}$$
Here $\mu(h(\boldsymbol{u}))$ and $\sigma(h(\boldsymbol{u}))$ denotes the h-dependent mean and standard deviation of the lognormal distribution, respectively. These are given by:
$\mu(h(\boldsymbol{u})) = a_0 + a_1 h ^ {a_2}$
$\sigma(h(\boldsymbol{u})) = b_0 + b_1 \exp(b_2 h)$
where $a_0 = 0.70, a_1 = 0.282, a_2 = 0.167, b_0 = 0.07, b_1 = 0.3449, b_2 = -0.2073$.
Furthermore, $\alpha = 1.76$ and $\beta = 1.59$.
