Confused about proof of Leibniz Integral Rule If we set $G(x) = \int_{0}^{x} f(x,y) dy$, then
\begin{align}
\frac{G(x+d)-G(x)}d =& \frac{\int_{0}^{x+d} f(x+d,y)dy - \int_{0}^{x} f(x,y)dy}d \\
=& \frac{\int_{0}^{x}f(x+d,y)dy+\int_{x}^{x+d} - \int_{0}^{x}f(x,y)dy}d
\end{align}
By grouping, $$\int_{0}^{x}\frac{f(x+d,y)-f(x,y)}ddy + \int_{x}^{x+d}f(x+d,y)dy,$$ which leads to:
$$\int_{0}^{x}f'(x,y)dy + \frac{\int_{0}^{x+d}f(x+d,y)dy-\int_{0}^{x}f(x+d,y)dy}d.$$
Why is the second term not equal to $f(x+d,x)$? I know it isn't, I'm just trying to see where I'm going wrong in the proof.

I think the main sticking point is understanding why 
$$\lim_{d\to 0}\frac{\int_{x}^{x+d}f(x+d,y)dy}{d}=f(x,x).$$
To prove this note that 
$$d\inf_{y\in[x,x+d]} f(x+d,y)\leq\int_{x}^{x+d}f(x+d,y)dy\leq d\sup_{y\in[x,x+d]} f(x+d,y),$$
assuming that $f$ is continuous on a compact set. Now simply divide through by $d$ and then take the limit as $d\to 0$. The squeeze theorem does the rest.
