Show that $\frac{d}{dx}\int_0^xf(x,y)dy = f(x,x) + \int_0^x \frac{\partial f}{\partial x}(x,y)dy$

Question

Here is the answer link, but I don't see one thing. Why is $$\frac{d}{dx}G(x,x) = \frac{ \partial G}{\partial u}(x,x) + \frac{\partial G}{\partial v}(x,x)$$ The way I understand this is that $G$ is a function of $u$ and $v$ that in this case take values $u=x, v=x$, so for me it looks like $\frac{d}{dx}$ is derivative with respect to value, like for example writing $f(z) = z^2$ and then asking what is $\frac{df}{d1}f(1)$. Or is this just treating $u, v$ as $u(x) = x, v(x) = x$?

Answer 1

Claim:

Let $\phi(x),\psi(x)\in C^1([\alpha,\beta]), a\lt \phi(x)\lt \psi(x)\lt b$ on $[\alpha,\beta]$. Assume that $f(x,y),f_x(x,y)$ are continuous on $[\alpha,\beta]\times [a,b]$, then$$\frac{d}{dx}\int_{\phi(x)}^{\psi(x)}f(x,y)dy=\int_{\phi(x)}^{\psi(x)}f_x(x,y)dy+f(x,\psi(x))\psi'(x)-f(x,\phi(x))\phi'(x)$$

Proof:

Let $g(x)=G(x,\phi(x)=u,\psi(x)=v)=\int_{\phi(x)}^{\psi(x)}f(x,y)dy$, then $g'(x)=G_x+G_{u}\phi'(x)+G_{v}\psi'(x)$. By the first fundamental theorem of calculus, $G_{v}=f(x,\psi(x))$ and $G_{u}=-f(x,\phi(x))$. Since $f_x(x,y)$ is continuous, $G_x=\int_{\phi(x)}^{\psi(x)}f_x(x,y)dy$ (this may also be proven more rigorously using $\epsilon-\delta$ argument). Combine the three parts and we have $$g'(x)=G_x+G_{u}\phi'(x)+G_{v}\psi'(x)=\int_{\phi(x)}^{\psi(x)}f_x(x,y)dy+f(x,\psi(x))\psi'(x)-f(x,\phi(x))\phi'(x)$$

Hence proven. Your problem is then the special case $\phi(x)=0,\psi(x)=x$, so

$$\int_0^x f(x,y)dy=\int_0^x f_x(x,y)dy+f(x,x)(1)-f(x,0)(0)=\int_0^x f_x(x,y)dy+f(x,x)$$