Conditional CDF given one dimension equals derivative of joint CDF towards that dimension divided by the density at that dimension?

Question

So I am familiar with the following:

$$P\left(X<x|Y=y\right) =\int_{-\infty}^{x}f\left(X=u|Y=y\right)du=\frac{1}{f\left(Y=y\right)}\cdot\int_{-\infty}^{x}f\left(X=u,Y=y\right)du$$

But during a lecture on copulas, I saw this:

$C_{U_2 \mid U_1}\left(u_2 \mid u_1\right)=\operatorname{Pr}\left(U_2 \leq u_2 \mid U_1=u_1\right)=\frac{\partial}{\partial u_1} C\left(u_1, u_2\right)$

(Because the marginals are standard uniform, you divide by 1 and thus can omit them).

But this means that you can write a conditional cdf as:

$$P\left(X<x|Y=y\right) = \frac{1}{f\left(Y=y\right)}\cdot\frac{\partial}{\partial y} F\left(x, y\right)$$

correct? As a copula is just a cdf.

Which would mean that $$\int_{-\infty}^{x}f\left(X=u,Y=y\right)du = \frac{\partial}{\partial y} F\left(x, y\right)$$

My question is why is the conditional cdf equal to the derivative of the joint cdf divided by the marginal? Or equivalently why: $$\int_{-\infty}^{x}f\left(X=u,Y=y\right)du = \frac{\partial}{\partial y} F\left(x, y\right)$$

Answer 1

First, the notations are somewhat confusing $$P\left(X<x|Y=y\right) =\int_{-\infty}^{x}f\left(X=u|Y=y\right)du=\frac{1}{f\left(Y=y\right)}\cdot\int_{-\infty}^{x}f\left(X=u,Y=y\right)du$$ should write $$ \mathbb P\left(X<x|Y\right) =\int_{-\infty}^{x}f_{X|Y}\left(u|Y\right)\text du=\frac{1}{f_Y\left(Y\right)}\cdot\int_{-\infty}^{x}f_{X,Y}\left(u,Y\right)\text du$$ since $f(Y=y)$ is inappropriate for a continuous rv.

Second, $$F(x,y)=\int_{-\infty}^{x}\int_{-\infty}^yf_{X,Y}\left(u,v\right)\text dv\text du$$ Hence, $$\frac{\partial}{\partial x}F(x,y)=\int_{-\infty}^yf_{X,Y}\left(x,v\right)\text dv\tag{1}$$ $$\frac{\partial}{\partial y}F(x,y)=\int_{-\infty}^xf_{X,Y}\left(u,y\right)\text du\tag{2}$$ and $$\frac{\partial^2}{\partial x\partial y}F(x,y)=f_{X,Y}(x,y)$$ The identity (2) can also write $$\frac{\partial}{\partial y}F(x,y)=\int_{-\infty}^xf_{X|Y}\left(u|y\right)f_Y(y)\text du=F_{X|Y}(x|y)f_Y(y)\tag{3}$$