I am reading through the book Probability and Random Processes for Electrical and Computer Engineers . On page 307, there is an equation which states the following.
With $F_{X|Y}$ representing the CDF of the conditional density $f(x | y)$, we are given the equation...
$$ F_Z(z) = \int_{-\sqrt{z}}^{\sqrt{z}} [F_{X|Y}(\sqrt{z - y^2} | y) - F_{X|Y}(-\sqrt{z - y^2} | y)]f_y(y) dy $$
The next line then introduced Leibniz' rule,
$$ \frac{d}{dz}\int_{a(z)}^{b(z)} h(z, y) dy = -h(z,a(z))a^\prime(z) + h(z,b(z))b^\prime(z) + \int_{a(z)}^{b(z)} \frac{\partial}{\partial z} h(z, y)dy $$
and then arrives at the conclusion on the next line that,
$$ f_Z(z) = \int_{-\sqrt{z}}^{\sqrt{z}} \frac{f_{X|Y(\sqrt{z - y^2} | y)} + f_{X|Y(-\sqrt{z - y^2} | y)}}{2\sqrt{z - y^2}} f_y(y) dy $$
but I cannot actually see how the steps are taken to get there. I am not even sure what the first step would be and how to apply Leibniz' rule. How do you apply Leibniz' rule here?
Too long for a comment.
The upper integration limit should not have a minus sign (book is correct here): $$ F_Z(z) = \int_{-\sqrt{z}}^{\sqrt{z}} \Big[F_{X|Y}\big(\sqrt{z - y^2} | y\big) - F_{X|Y}\big(-\sqrt{z - y^2} | y\big)\Big]f_y(y)\,dy\,. $$ This is a difference of two integrals of the form $$ G(z)=\int_0^{\pm\sqrt{z}}g\big(\sqrt{z-y^2}\big)\,f_y(y)\,dy\,. $$ In your version of the Leibniz rule you have written $h(z,y)=g\big(\sqrt{z-y^2}\big)\,f_y(y)$. Can you proceed ?