I've been wondering if there is a fairly simple proof or derivation for the following (called the Leibniz Rule):
$$ \frac{d}{dt} \int_{a(t)}^{b(t)}f(x,t)dx = \int_{a(t)}^{b(t)}\frac{\partial{f}}{\partial{t}}dx+ f(b(t),t)\frac{d}{dt}b(t)-f(a(t),t)\frac{d}{dt}a(t)$$
What I tried doing is integrating by parts the leftmost integral, and then applying the time derivative to what it's equal to. However, it seems the other side is very far from what the result above. A small hint would suffice as I like trying to prove things on my own.
Best Answer
HINT: first try to see what is $\partial _y\int_{a}^y f(x,t)\mathop{}\!d x$ and $\partial _t \int_{a}^y f(x,t)\mathop{}\!d x$, the first case follows from the fundamental theorem of calculus, the latter from the continuity of $\partial_t f$ and the definition of partial derivative.
Then for $F(y,t):=\int_{a}^y f(x,t)\mathop{}\!d x$ you will get an expression for $\nabla F(y,t)$, then setting $\varphi (t):=(g(t),t)$ for some differentiable $g$ you can compute $(F\circ \varphi )'(t)$ using the chain rule.
Finally the general case follows from the decomposition $$ \int_{h(t)}^{g(t)}f(x,t)\mathop{}\!d x=\int_{a}^{g(t)}f(x,t)\mathop{}\!d x-\int_{a}^{h(t)}f(x,t)\mathop{}\!d x $$
We will show that $\partial _t \int_{a}^{y}f(x,t)\mathop{}\!d x=\int_{a}^y \partial _t f(x,t)\mathop{}\!d x$ when $\partial _t f$ is continuous. Note that, as the partial derivative in $t$ of $f$ is continuous then for fixed $t_0$ the function defined by
$$ q_{t_0}(x,h):=\begin{cases} \frac{f(x,t_0+h)-f(x,t_0)}{h},&h\neq 0\\ \partial _t f(x,t_0),& h=0 \end{cases} $$
is continuous, therefore it is uniformly continuous in the compact set $[a,y]\times [-r ,r ]$ for any chosen $r >0$, so for every $\epsilon >0$ there is some $\delta \in(0,r)$ such that $|q_{t_0}(x,h)-\partial _t f(x,t_0)|<\epsilon $ for all $(x,h)\in[a,y]\times (-\delta ,\delta )$, therefore
$$ \left|\int_{a}^y (q_{t_0}(x,h)-\partial _t f(x,t_0))\mathop{}\!d x\right|\leqslant \int_{a}^y |q_{t_0}(x,h)-\partial _t f(x,t_0)|\mathop{}\!d x<(y-a)\epsilon $$
what shows that
$$ \partial _t \int_{a}^y f(x,t_0)\mathop{}\!d x=\lim_{h\to 0}\int_{a}^y q_{t_0}(x,h)\mathop{}\!d x=\int_{a}^y \partial _t f(x,t_0)\mathop{}\!d x $$