Formula for derivative of $F(x)=\int_{\phi(x)}^{\psi(x)} f(x,t) dt$ [duplicate]

Let $I$ and $J$ be intervalls, $\phi:I\rightarrow J$ and $\psi:I\rightarrow J$ differentiable functions and $f:I\times J\rightarrow \mathbb{R}$ continuously differentiable twice (in $C^2$).

Find a formula for the derivative of $F(x)=\int_{\phi(x)}^{\psi(x)} f(x,t) dt$ with $x\in I$. Hint: Chain rule

My attempt:

I did following: We know that $\int_a^b f(x) dx = F(b)-F(a)$. So therefore, \begin{align*} \frac{dF(x)}{dx}&=\frac{d [F(\psi(x))-F(\phi(x))]}{dx}=f(\psi(x))\cdot \frac{d\psi(x)}{dx}-f(\phi(x))\cdot \frac{d\phi(x)}{dx}\\ &=f(\psi(x))\cdot\psi'(x)-f(\phi(x))\cdot \phi'(x) \end{align*} I'm pretty sure that's right for $\int_{\phi(x)}^{\psi(x)} f(t) dt$, but I'm not sure if that's correct for $f(x,t)$. Any remarks?

Let $a\in J$ and $g(x,t) = \int_a^t f(x,t)$. Then, $\partial_2 g = f$ and : $$\partial_1 g(x,t) = \int_a^t \partial_1f(x,t)\text dt$$ This can be checked because, $\partial_2f(x,t)$ is bounded on a compact neighborhood of $\{x\} \times [a,t]$ in $I\times J$. Then, we have, by the fundamental theorem of calculus :

$$F(x) = g(x,\psi(x)) - g(x,\phi (x))$$ and, by the chain rule : \begin{align} F'(x) &= \int_a^{\psi(x)}\partial_1 f(x,t)\text dt + f(x,\psi(x)) - \int_a^{\phi(x)}\partial_1 f(x,t)\text dt - f(x,\phi(x)) \\ &= f(x,\psi(x)) -f(x,\phi(x)) + \int_{\phi(x)}^{\psi(x)} \partial_1f(x,t)\text dt \end{align}