While trying to prove something, I came up with a strange case involving the use of Leibniz integral rule.
Let a variable $x$ be parametrized by $t$ so that $x=x(t)$.
Now I want to take the derivative with respect to $x(t)$ an integral:
$$\frac{d}{dx(t)}\int_0^tf(x(t'))dt'$$ for a well-behaved function $f(x(t))$ and with $\frac{d}{dx(t)}$ representing just the normal derivative wrt $x(t)$.
The problem is that I don't know how to actually apply the Leibniz rule to the above operator.
Any help is appreciated.
EDIT: Also, how does it generalize when $x(t)\rightarrow\vec{x(t)}=(x(t), y(t), z(t))$ and $\frac{d}{dx(t)}\rightarrow\nabla$?
What about this:
$$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x} \int_0^t \mathrm{d}t' f(x(t'))&=&\frac{\mathrm{d}}{\mathrm{d} x} \frac{\mathrm{d}t}{\mathrm{d} t}\int_0^t \mathrm{d}t' f(x(t'))\\ &=&\frac{1}{ \frac{\mathrm{d}x}{\mathrm{d}t}}\frac{\mathrm{d}}{\mathrm{d}t}\int_0^t \mathrm{d}t' g(t')\\ &=&\frac{1}{ \frac{\mathrm{d}x}{\mathrm{d}t}} g(t)\\ &=&\frac{1}{ \frac{\mathrm{d}x}{\mathrm{d}t}} f(x(t)) \end{align} $$