I don't think so. A joint distribution has domain $(-\infty, \infty) \times (-\infty, \infty)$. If we partition each component of the cartesian product in two by selecting some value $x$ and some value $y$, then we get $4$ subsets,
$$(-\infty, x] \times (-\infty, y],\;\;(-\infty, x] \times [y,\infty),\\ [x, \infty) \times (-\infty, y],\;\;[x, \infty) \times [y,\infty)$$
Somade up of intersections of two events,
$$\begin{align}\Pr(X\le x, Y\le y) =& 1- \Pr(X\le x, Y\ge y)\\ &-\Pr(X\ge x, Y\le y)\\ &-\Pr(X\ge x, Y\ge y) \end{align}$$$$A = P(X\le x), \;\; B = P(Y\le y)$$
and you should differentiate all three cdf'stheir corresponding complements.
Then (as the OP noted in a commnent),
$$\Pr(X\ge x, Y\ge y) = P(A^c\cap B^c) = 1 - P(A\cup B)$$
$$=1-\big[P(A) + P(B) - P(A\cap B)\big]$$
So it appears that by taking the rightcross-hand side. In general, There is nothing "special" aboutpartial derivative of $\Pr(X\ge x, Y\ge y)$ we should again get the joint density. Let's verify that would permit you to derive:
$$\Pr(X\ge x, Y\ge y) = \int_x^{\infty}\int_y^{\infty}f(s,t)dtds$$
$$\frac {\partial \Pr(X\ge x, Y\ge y)}{\partial y} = \int_x^{\infty} \left(\frac{\partial}{\partial y}\int_y^{\infty}f(s,t)dt\right)ds $$
$$=\int_x^{\infty}-f(s,y) ds$$
$$\frac {\partial^2 \Pr(X\ge x, Y\ge y)}{\partial y\partial x} = \frac {\partial }{\partial x} \int_x^{\infty}-f(s,y) ds = -\left(-f(x,y)\right) = f(x,y)$$
The above also means that we can obtain the joint pdf from itany of the four joint events indicated by the breakdown of the support -but in the other two cases, we should multiply by $-1$.
$$\begin{align} f(x,y) =& \frac {\partial^2 \Pr(X\le x, Y\le y)}{\partial y\partial x}\\ =&\frac {\partial^2 \Pr(X\ge x, Y\ge y)}{\partial y\partial x}\\ =&-\frac {\partial^2 \Pr(X\le x, Y\ge y)}{\partial y\partial x}\\ =&-\frac {\partial^2 \Pr(X\ge x, Y\le y)}{\partial y\partial x} \end{align}$$