Problem: Let $h: \mathbb{R}^2 \to \mathbb{R}$ be continuous and differentiable with respect to its second variable, define $u(t)= \displaystyle \int_0^t h(s,t)ds, \ t \in \mathbb{R}$
In an exercise before I already had to show that the above function is differentiable, meaning with respect to $x$ and $y$ by defining the function $f(x,y)= \int_0^x h(s,y)ds$
Now I am supposed to compute $u'(t)$ and I struggle still a lot with the multivariable chain rule.
Theorem (Multivariable Chainrule): Let $f : \Omega \subset \mathbb{R}^n \to \mathbb{R}^m $ and $g: U \subset \mathbb{R}^m \to \mathbb{R}^l$ be differentiable functions. The composition $g \circ f$ is differentiable with derivative $$d(g \circ f ) (x) = dg(f(x)) \cdot df(x) $$
I know that the dot "$\cdot$" in the theorem actually stand for Matrix multiplication, which makes intuitive sense to me since we're working in $\mathbb{R}^n$. However I can't really grasp the context yet.
My (intuitive) approach: In the exercise on the whether or not $u$ is differentiable I had to define $ f(x,y)= \int_0^x h(s,y)ds$ and I managed to show that $$ \partial_yf = \int_0^x \partial_yh(s,y)ds \\ \partial_xf = h(z,y) $$ and I am not quite sure how this is supposed to help me with finding the derivative of $u$. Following my still dominant instinct from single variable Calculus I would just have said that $$u'(t)= h(t,t) \cdot \partial_th(s,t)ds $$
Addition: I also thought about defining the below composition: \begin{align} \begin{cases} \begin{matrix} \mathbb{R}^2 & \overset{h}{\longrightarrow} &\mathbb{R} & \overset{p}{\longrightarrow}& \mathbb{R} \\ (s,y) & \longmapsto &h(s,y) \\ & & h(s,y) & \longmapsto & \displaystyle \int_0^t h(s,t)ds \end{matrix} \end{cases} \end{align} which might help me more to understand what's going on. I would appreciate one or two paragraphs that might help me to grasp the concept of the multivariable chain rule.
If $u(t)=f(t,t)$ then $u'(t)=\dfrac{\partial f}{\partial x}(t,t)+\dfrac{\partial f}{\partial y}(t,t)$.
Edit: This can be extended into $u=f\circ(g_1,g_2)$ with $f$ as in the question and $g_1=g_2:\mathbb R\to\mathbb R$ defined by $g_1(t)=g_2(t)=t$. Thus, $$ u'=\partial_xf(g_1,g_2)\cdot g'_1+\partial_yf(g_1,g_2)\cdot g'_2. $$