Proving that $F(x)=\int_{a}^{x}f(x,y)dy$ is continuously differentiable

I know that $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is continuous, $f$ is differentiable in $x$ with $D_{1}f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ continuous and I know that: $F(x)=\int_{a}^{x}f(x,y)dy, x\in \mathbb{R}$ .

I now want to prove that $F$ is continously differentiable (thus that $F$ is differentiable and that $F'(x)$ is continuous) I also need to show that: $$F'(x)=f(x,x)+\int_{a}^{x}\frac{\partial f(x,y)}{\partial x}dy, x \in \mathbb{R}$$.
I know that there exists a $G$ such that: $\int_{a}^{x}f(x,y)dy=G(x)-G(a)$ with $G'(x)=f(x,y)$, and that $\frac{d}{dx}\int_{a}^{b}f(x,y)dy=\int_{a}^{b}\frac{\partial f(x,y)}{\partial x}dy$, that's the furthest I got with this problem can anyone help me solve this?

• So you allow $f(x,y)$ is differentiable with respect to $x$, but that derivative is not jointly continuous? So you will even have to prove that $\int_{a}^{x}\frac{\partial f(x,y)}{\partial x}dy$ exists? Sep 25, 2017 at 15:03
• Is $D_1f$ continuous as well?
– edm
Sep 25, 2017 at 15:18
• Yes, $D_1f$ is continuous Sep 25, 2017 at 15:23

You need the chain rule for 2 dimensions (see https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions). Here is what I mean: Define $g:\mathbb{R}\rightarrow\mathbb{R}^2$ by $$g(x)=(x,x)$$ and $h:\mathbb{R}^2\rightarrow\mathbb{R}$ by $$h(x,y)=\int_a^x f(y,z)\, dz.$$ Now note that $$F(x)=h(g(x))$$ and use chain rule.