Special case of Leibniz formula I am wondering if I understand it right. Considering the following partial derivative:
$$
\partial_tF(t, x) = \partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds
$$
It looks to me as Leibniz should be used. So we should get
$$
\partial_t\int_{a(t)}^{b(t)} f(t-s, x)ds \\ = \int_{a(t)}^{b(t)} \partial_t f(t-s, x)ds - f(t-a(t), x)(1-a'(t)) + f(t-b(t), x)(1-b'(t))
$$
Is this correct, or did I get something wrong?
 A: 
Applying the Leibniz integral rule we obtain
\begin{align*}
\color{blue}{\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f\left(u(t,s),x\right)\,ds\right)}
&\color{blue}{=f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)}\\
&\qquad\color{blue}{-f(u(t,a(t)),x)\,\frac{d}{dt}a(t)}\\
&\qquad\color{blue}{+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds}\tag{1}
\end{align*}
provided the functions are appropriately differentiable.

Example: We look at a small example.
\begin{align*}
u(t,s)&=t-s\qquad\qquad a(t)=4t\qquad b(t)=3t^2\\
f(u(t,s),x)&=u(t,s) x\\
&=(t-s)x
\end{align*}

*

*The left-hand side of (1) gives
\begin{align*}
\frac{d}{dt}\left(\int_{a(t)}^{b(t)}f(u(t,s),x)\,ds\right)&=\frac{d}{dt}\int_{s=4t}^{3t^2}(t-s)x\,ds\\
&=x\frac{d}{dt}\left(ts-\frac{1}{2}s^2\bigg|_{s=4t}^{3t^2}\right)\\
&=x\frac{d}{dt}\left(3t^3-\frac{9}{2}t^4-4t^2+8t^2\right)\\
&\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)}
\end{align*}

*The right-hand side of (1) gives
\begin{align*}
&f\left(u(t,b(t)),x\right)\,\frac{d}{dt}b(t)-f(u(t,a(t)),x)\,\frac{d}{dt}a(t)
+\int_{a(t)}^{b(t)}\frac{\partial}{\partial t}f\left(u(t,s),x\right)\,ds\\
&\qquad=\left(t-\left(3t^2\right)\right)x\cdot 6t-\left(t-(4t)\right)x\cdot 4
+\int_{4t}^{3t^2}\frac{d}{dt}(t-s)x\,ds\\
&\qquad=\left(6t^2x-18t^3x\right)+12tx+x\int_{4t}^{3t^2}\,ds\\
&\qquad=\left(6t^2-18t^3\right)x+12tx+\left(3t^2-4t\right)x\\
&\qquad\,\,\color{blue}{=x\left(-18t^3+9t^2+8t\right)}
\end{align*}
and we see according to (1) the blue marked results coincide as expected.

