Find $f'$ for the function $f(x,y) = \int_{a}^{x + y}g$ Find $f'$ for the function $f(x,y) = \int_{a}^{x + y}g$, where $g:\mathbb{R} \to \mathbb{R}$ is continuous.
My question has to do more with the concept than solving this specific expression. How do I interpret the composition? This exercise appears in my class with regards to applying the chain rule.  I remember that it has to do with the fundamental theorem of calculus. Mechanically I know that the answer will be:
$$f(x,y) = \int_{a}^{x + y}g(z) dz \Rightarrow f'(x,y) = g(x+y) \cdot [1,1]$$
but I'm not seeing how the composition of the functions comes about. Is it an expression of the sort: 
$$ F(g(z)) =  \int_{a}^{x+y}g(z),\  \text{where}\ F = \int_{a}^{x + y}\phi$$
As you can see, all the symbols and variables are confusing me a bit. Could somebody give me a bit of an explanation?
Edit: Possible Interpretation
One of the things I have been working on understanding is performing the differentiation process in matrix/ vector notation. With that being said this expression is a composition of the functions:
$$F(u) =\int_{a}^{u}g(t)dt\ \text{where} \ F:\mathbb{R} \to \mathbb{R}\ \text{and} \\ \ u(x,y) = x + y, \ \text{where}\ u: \mathbb{R}^{2} \to \mathbb{R} \\ \Rightarrow (F \circ u)(x,y) = F(u(x,y))$$
So when the question asks for $f'$, what is being asked for is 
$$F'(u) = DF(u(x,y)) \cdot Du(x,y) \\  DF(u(x,y)) = [D_{1}F] = [g(u)] \\ Du(x,y) = [D_{x}u(x,y), D_{y}u(x,y)] = [1,1] \\ \therefore \ DF(u(x,y)) \cdot Du(x,y) = g(u) \cdot [1,1] \\ = [g(u), g(u)] \ \text{since} \ u = x + y \\ = [g(x+y), g(x+y)] $$
From this what I have to be aware of is that the composition involves the integrand and one of the limits of integration. The function $g(t)$ is just a part of the integrand function.
 A: The differential of $f$ is $g(x+y)dx+g(x+y)dy$
A: Your expression is of the form $f(x,y)=F(x+y)$, where $F(z)=\int_0^z g(t)\,dt$. The "chain rule" you need to apply here is the one-variable one. So
$$
f'(x,y)=[f_x,f_y]=[g(x+y),g(x+y)].
$$
In both partial derivatives you do $\tfrac{\partial}{\partial x}(F(x+y))=F'(x+y)\,\tfrac{\partial}{\partial x}(x+y)=F'(x+y)=g(x+y)$.
A: You're give a continuous $g: \Bbb{R} \to \Bbb{R}$ and a function $f: \Bbb{R}^2 \to \Bbb{R}$ defined by $f(x,y) = \int_a^{x+y}g$. Here's the precise composition going on. First, define
$G: \Bbb{R} \to \Bbb{R}$ by $G(\xi ) = \int_a^\xi g$, and define $\sigma: \Bbb{R}^2 \to \Bbb{R}$ by $\sigma(x,y) = x+y$. Then, as an equation involving functions alone, you can write that
\begin{align}
f = G \circ \sigma
\end{align}
Now, by the standard FTC of single variable calculus, since $g$ is continuous everywhere, it follows that $G$ is differentiable and that $G' = g$. Next, $\sigma: \Bbb{R}^2 \to \Bbb{R}$ is a linear transformation and hence trivially is differentiable, and the matrix representation of its derivative is its own matrix representation. Thus, by the (multivariable) chain rule it follows that $f$ is also differentiable, and that for all $(x,y) \in \Bbb{R}^2$,
\begin{align}
f'(x,y) &= (G\circ \sigma)'(x,y) \\
&= G'(\sigma(x,y)) \cdot \sigma'(x,y) \\
&= g(\sigma(x,y)) \cdot [\sigma] \\
&= g(x+y) \cdot 
\begin{bmatrix}
1 & 1
\end{bmatrix}.
\end{align}
(here I used $[\sigma]$ to mean the matrix representation of the linear transformation $\sigma$ relative to the standard basis of $\Bbb{R}^2$).

The usual way of computing $f'(x,y)$ is to simply compute its partial derivatives, and put them into a matrix. If you prefer this approach then go for it, but be sure you're able to justify why $f$ is in fact differentiable (because mere existence of partial derivatives doesn't guarantee differentiability).
A: Just consider the composition:
\begin{alignat}{2}
\mathbf R\times\mathbf R&\xrightarrow{\enspace\:s\enspace\:}\enspace \mathbf R\xrightarrow{\enspace\:g \quad\:}\mathbf R\\
(x,y)&\xrightarrow{\phantom{\enspace\:s\enspace\:}}x+y\xrightarrow{\hphantom{\enspace g \enspace}}g(x+y)
\end{alignat}
A: Suppose the definition domain of $f$ is an open $\Omega$ in $\mathbb{R}^2$. Let $\mathcal{L}(\mathbb{R}^2;\mathbb{R})$ the set of all linear transformation $T:\mathbb{R}^2\to\mathbb{R}$. The derivative $f^\prime$ is a aplication 
$$
\begin{array}{cccc}
f^{\prime}: & \Omega & \longrightarrow & \mathscr{L}(\mathbb{R}^2,\mathbb{R})\\
            &   (x,y)   & \longmapsto     & f^\prime(x,y)
\end{array}
$$
and the $f^\prime(x_0,y_0)$ is the linear application 
$$
f^\prime(x_0,y_0)\cdot(u,v)=
\left\langle 
\left(\frac{\partial f(x_0,y_0)}{\partial x}, \frac{\partial f(x_0,y_0)}{\partial y}\right)\,,\,(u,v)
\right\rangle
=
\frac{f(x_0,y_0)}{\partial x}\cdot u+\frac{\partial f(x_0,y_0)}{\partial y}\cdot v
$$
We have 
\begin{align}
\frac{f(x_0,y_0)}{\partial x}
=&
\lim_{u\to 0}\frac{\int_{a}^{(x_0+u)+y_0}g(t)\mathrm{d}t-\int_{a}^{x_0+y_0}g(t)\mathrm{d}t}{u}
\\
=&
\lim_{u\to 0}\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}g(t)\mathrm{d}t}{u}
\\
=&
\lim_{u\to 0}\left[\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}g(t)-g(x_0+y_0)\mathrm{d}t}{u}\right]
\\
&\hspace{2cm}
+
\lim_{u\to 0}\left[\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}g(x_0+y_0)\mathrm{d}t}{u}\right]
\\
=&
\lim_{u\to 0}\left[\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}g(t)-g(x_0+y_0)\mathrm{d}t}{u}\right]
+g(x_0+y_0)
\\
\end{align}
Inside of integral sinal we have $t\in (x_0+y_0-|u|,x_0+y_0+|u|)$ and 
\begin{align}
\left|\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}g(t)-g(x_0+y_0)\mathrm{d}t}{u}\right|
\leq &
\frac{\int_{x_0+y_0}^{(x_0+u)+y_0}|g(t)-g(x_0+y_0)|\mathrm{d}t}{|u|}
\\
\leq & 
\frac{\sup_{t}|g(t)-g(x_0+y_0)|\cdot |u|}{|u|}
\\
&=
\sup_{t}|g(t)-g(x_0+y_0)|
\end{align} 
Making $u\to 0$, we have $\sup_{t}|g(t)-g(x_0+y_0)|\to 0$. And therefore 
$$
\frac{f(x_0,y_0)}{\partial x}=g(x_0+y_0)
$$
In an entirely analogous way we have
$$
\frac{f(x_0,y_0)}{\partial y}=g(x_0+y_0)
$$
