$$
\begin{align}
\frac{\partial y}{\partial \gamma } =\;&
e^{\lambda \, t} \, x_1^{\nu \, \delta} \; x_2^{\nu \, ( 1 - \delta )}
\exp \left( - \frac{\rho}{2} \, \nu \, \delta \, ( 1 - \delta )
\left( \ln x_1 - \ln x_2 \right)^2
\right)
\label{eq:derivYGammaApprox}\\
\frac{\partial y}{\partial \delta } =\;&
\gamma \, e^{\lambda \, t} \, \nu \, \left( \ln x_1 - \ln x_2 \right)
x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
\label{eq:derivYDeltaApprox}\\
& \left( 1 - \frac{ \rho }{ 2 }
\big[ 1 - 2 \, \delta + \nu \, \delta ( 1 - \delta )
\left( \ln x_1 - \ln x_2 \right) \big]
\left( \ln x_1 - \ln x_2 \right)
\right)
\nonumber\\
\frac{\partial y}{\partial \rho } =\;&
\gamma \, e^{\lambda \, t} \, \nu \, \delta \, ( 1 - \delta ) \,
x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
\bigg( - \frac{1}{2} \left( \ln x_1 - \ln x_2 \right)^2
\label{eq:derivYRhoApprox}\\
& + \frac{\rho}{3} ( 1 - 2 \, \delta ) \left( \ln x_1 - \ln x_2 \right)^3
+ \frac{\rho}{4} \, \nu \, \delta ( 1 - \delta )
\left( \ln x_1 - \ln x_2 \right)^4
\bigg)
\nonumber\\
\frac{\partial y}{\partial \nu } =\;&
\gamma \, e^{\lambda \, t} \, x_1^{ \nu \, \delta } \; x_2^{ \nu ( 1 - \delta ) }
\bigg( \delta \ln x_1 + ( 1 - \delta ) \ln x_2
\label{eq:derivYNuApprox}\\
& - \frac{\rho}{2} \, \delta ( 1 - \delta )
\left( \ln x_1 - \ln x_2 \right)^2
\left[ 1 + \nu
\left( \delta \ln x_1 + ( 1 - \delta ) \ln x_2 \right)
\right]
\bigg)
\nonumber
\end{align}
$$
If $\rho$ is zero or close to zero,
the partial derivatives with respect to $\lambda$ are calculated
also with Equation~\ref{eq:derivYLambda},
but now $\partial y / \partial \gamma$ is calculated
with Equation