It is known that $$E[X\vert Y\geq y](1-F(y)) = \int_y^\infty E[X\vert Y=t]f(t)dt$$
To take the derivative, I think this is just Leibniz rule, but I don't know how to handle the conditional expectation on the RHS in the integral.
Do I need to write the conditional Expectation as an integral and then take the derivative of the double integral?
I don't see double integrals in the answer though, which is why I am confused. The answer is
$$\frac{d}{dy}E[X\vert Y\geq y] = f(y)\frac{E[X\vert Y\geq y] - E[X\vert Y=y]}{1-F(y)}$$
Also, if I use Leibniz rule I don't get an $E[X\vert Y\geq y ]$ term
We just need to differentiate $\int_y^\infty f(\tau) \,d\tau$: $$\frac d {dy} \operatorname{E}(X \mid Y \geq y) = \frac d {dy} \frac {\int_y^\infty \operatorname{E}(X \mid Y = \tau) \,f_Y(\tau) \,d\tau} {\operatorname{P}(Y \geq y)} = \\ -\operatorname{E}(X \mid Y \geq y) \,\frac {\frac d {dy} \operatorname{P}(Y \geq y)} {\operatorname{P}(Y \geq y)} + \frac {\frac d {dy} \int_y^\infty \operatorname{E}(X \mid Y = \tau) \,f_Y(\tau) \,d\tau} {\operatorname{P}(Y \geq y)} = \\ \operatorname{E}(X \mid Y \geq y) \,\frac {f_Y(y)} {1 - F_Y(y)} - \frac {\operatorname{E}(X \mid Y = y)\,f_Y(y)} {1 - F_Y(y)}.$$