How to compute the seconds-order partial derivatives of Gaussion copula? How to compute the following seconds-order  partial derivatives of u and v for Gaussion copula?Thanks

 A: A copula is a multivariate distribution function with uniform marginals.  It is used to impose or alter the dependence structure of a set of random variables with known marginal distributions. The density of a bivariate copula is the mixed second partial derivative.
Differentiate with respect to $u$ using Leibniz's rule.
$$\frac{\partial}{\partial u}C(u,v;\rho)\\=\frac{d}{du}\Phi^{-1}(u)\frac{1}{2\pi \sqrt{1-\rho^2}}\int_{-\infty}^{\Phi^{-1}(v)}\exp\left[-\frac1{2}\frac{y^2-2\rho\Phi^{-1}(u)y+\Phi^{-1}(u)^2}{1-\rho^2}\right]\,dy.$$
Do the integration directly by completing a square.
$$\int_{-\infty}^{\Phi^{-1}(v)}\exp\left[-\frac1{2}\frac{y^2-2\rho\Phi^{-1}(u)y+\Phi^{-1}(u)^2}{1-\rho^2}\right]\,dy\\=\int_{-\infty}^{\Phi^{-1}(v)}\exp\left[-\frac1{2}\frac{[y-\rho\Phi^{-1}(u)]^2+(1-\rho^2)\Phi^{-1}(u)^2}{1-\rho^2}\right]\,dy.$$
Then 
$$\frac{\partial}{\partial u}C(u,v;\rho)=\frac{d}{du}\Phi^{-1}(u)\frac{1}{\sqrt{2\pi} }\exp\left[-\frac{\Phi^{-1}(u)^2}{2}\right]\Phi\left[\frac{\Phi^{-1}(v)-\rho\Phi^{-1}(u)}{\sqrt{1-\rho^2}}\right].$$
The derivative of the inverse standard normal distribution is
$$\frac{d}{du}\Phi^{-1}(u)=\frac1{\Phi'(\Phi^{-1}(u))}=\frac1{\frac{1}{\sqrt{2\pi} }\exp\left[-\frac{\Phi^{-1}(u)^2}{2}\right]}.$$
Hence
$$\frac{\partial}{\partial u}C(u,v;\rho)=\Phi\left[\frac{\Phi^{-1}(v)-\rho\Phi^{-1}(u)}{\sqrt{1-\rho^2}}\right].$$
Now differentiate with respect to $v$ -- again using the above formula for the derivative of the inverse standard normal distribution function.
$$\frac{\partial^2}{\partial u \partial v}C(u,v;\rho)=\frac1{\sqrt{1-\rho^2}}\frac{d}{dv}\Phi^{-1}(v)\Phi'\left[\frac{\Phi^{-1}(v)-\rho\Phi^{-1}(u)}{\sqrt{1-\rho^2}}\right]\\=\frac1{\sqrt{1-\rho^2}}\left[\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{\Phi^{-1}(v)^2}{2}\right)\right]^{-1}\frac{1}{\sqrt{2\pi} }\exp\left[\frac{-\Phi^{-1}(v)^2+2\rho\Phi^{-1}(u)\Phi^{-1}(v)-\rho^2\Phi^{-1}(u)^2}{2(1-\rho^2)}\right]\\=\frac1{\sqrt{1-\rho^2}}\exp\left[\frac{-\rho^2\Phi^{-1}(v)^2+2\rho\Phi^{-1}(u)\Phi^{-1}(v)-\rho^2\Phi^{-1}(u)^2}{2(1-\rho^2)}\right].$$
