Tail dependence index for Gaussian copula is 0 Why is Gaussian Copula's Tail Dependence Zero?
I am confused about the second equation. Why does the derivative of C(q,q) can be written in two parts? And why each part has a conditional probability form? Is it due to the full derivative?

 A: I think maybe it can solve as follow:
By definition, 
And $$
\frac{\partial C(U,V)}{\partial x} = \frac{\partial C(U,V)}{\partial U}\frac{\partial U}{\partial x} + \frac{\partial C(U,V)}{\partial V}\frac{\partial V}{\partial x}
$$
Since,
 $$
U\sim U(0,1)
$$
$$
U(x)=P(U\leq x)=x
$$
Then,
$$
\frac{\partial U}{\partial x}=\frac{\partial x}{\partial x}=1
$$
Similar to V, I have:
$$
\frac{\partial C(U,V)}{\partial x} = \frac{\partial C(U,V)}{\partial U} + \frac{\partial C(U,V)}{\partial V}
$$
By the symmetry property of copula,
$$
\frac{\partial C(U,V)}{\partial x} = 2\frac{\partial C(U,V)}{\partial V}
$$
Then, I have to show:
$$
\frac{\partial C(U,V)}{\partial V}=C_{1|2}(u|v)
$$
where 
$$
C_{1|2}(u|v) = P(U\leq u|V=v)
$$
Consider the conditional density of V given U=u is:
$$
h_{1|2}(u|v)=\frac{h(u,v)}{f_{V}(v)}=h(u,v)
$$
Then, on the one side,
$$
C_{1|2}(u|v)=P(U\leq u|V=v)=\begin{matrix} \int_{0}^{u} h_{1|2}(x|v)\, dx \end{matrix}=\begin{matrix} \int_{0}^{u} h(x,v)\, dx \end{matrix}
$$
On the other side,
$$
C(u,v)=\int_{0}^{v} \int_{0}^{u}h(x,y) \,dy\,dx
$$
so that:
$$
\frac{\partial C(U,V)}{\partial V} = \begin{matrix} \int_{0}^{u} h(x,v)\, dx \end{matrix}
$$
Thus, I get:
$$
\frac{\partial C(U,V)}{\partial V}=C_{1|2}(u|v) = P(U\leq u|V=v)
$$
To sum up, I can show that:
$$
\frac{\partial C(q,q)}{\partial q} = P(U_2 \leq q|U_1 =q)+P(U_1 \leq q|U_2 =q)
$$
And take the limit, equation in my question will be presented.
I am not sure whether it is correct or not. Let's discuss. Thx.
