I did not get the first step of the derivation in this question
The correct is: \begin{align}\frac{d}{dx} \int_{0}^{x} dy \left[\int_{0}^{x} f(y,z)\, dz \right] &=\int_{0}^{x} f(x,z) \,dz + \int_{0}^{x} dy\, \frac{\partial}{\partial x}\left[\int_{0}^{x} f(y,z)\, dz \right]. \end{align}
But, for me it should be (using the derivative product rule):
\begin{align}\frac{d}{dx} \int_{0}^{x} dy \left[\int_{0}^{x} f(y,z)\, dz \right] &=\frac{d}{dx} \left(\int_{0}^{x} dy \right)\int_{0}^{x} f(y,z)\, dz + \int_{0}^{x} dy \,\frac{d}{dx} \left(\int_{0}^{x} f(y,z)\, dz \right)\\ &=\int_{0}^{x} f(y,z)\, dz + \int_{0}^{x} dy\, \frac{\partial}{\partial x}\left[\int_{0}^{x} f(y,z)\, dz \right]. \end{align}
So I am putting a $y$ in place of an $x$.
Could you please detail this first step and explain why I am wrong?
Substitute $F(x,y) = \int_0^x f(y,z) dz$. Then we want to evaluate $$\frac{d}{dx}\int_0^xF(x,y) dy.$$ By Leibniz rule, \begin{align} \frac{d}{dx}\int_0^xF(x,y) dy &= F(x,x) + \int_0^x \frac{\partial}{\partial x}F(x,y)dy \\ &= \int_0^xf(x,z)dz + \int_0^x \frac{\partial}{\partial x}\left[\int_{0}^{x} f(y,z)\, dz \right] dy \end{align} which is what was found in the other post.