Find $\frac{d}{dx} \left ( \int_{a(x)}^{b(x)} f(x,t) dt \right )$

Question

I'm trying to show that$$\frac{d}{dx} \left ( \int_{a(x)}^{b(x)} f(x,t) dt \right )=\int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(x,t)dt +b'(x)f(x,b(x))-a'(x)f(x,a(x))$$ Such that $f$ and $\frac{\partial f}{\partial x}$ are continuous on a rectangle $I \times J$, and $a,b: I \to J$ are continuously differentiable.
Define $\phi$ on $I \times J \times J$ by: $$\phi(x,y,z)=\int_y^z f(x,t) dt$$ Hence the we are looking to differentiate the function $x \mapsto\phi(x,a(x),b(x))$. In order to apply the Chain Rule I will need to verify that $$\frac{\partial \phi}{\partial x}(x,y,z)=\int_y^z \frac{\partial f}{\partial x}(x,t)dt$$ Which is true if we restrict $x$ to a compact interval, by the Leibniz Rule. This would make the argument "local", and I would have to define a compact neighborhood of $x$ in the beginning (I'm not sure about this part).
The other two term in the Chain Rule formula are a result of the Fundamental Theorem. I think my reasoning is correct but my book hints proving that $(x,y,z) \mapsto \int_y^z \frac{\partial f}{\partial x}(x,t)dt$ is continuous, and I can't tell if I need this or not.

Answer 1

I made the mistake off applying the chain rule to $\phi$ without verifying that it's a differentiable function. So I will prove that, by showing that its partial derivatives are continuous on a particular parallelepiped in $I \times J \times J$. Let $I'$ and $J'$ be compact intervals of $I$ and $J$ respectively. The partial derivatives of $\phi$ w.r.t $y$ and $z$ are continuous on $I \times J \times J$ by the assumptions on $f$.

For all $(x,y,z)$ and $(x_0,y_0,z_0)$ in $I' \times J' \times J'$: $$\left | \frac{\partial \phi}{\partial x}(x,y,z)-\frac{\partial \phi}{\partial x}(x_0,y_0,z_0) \right |=\left |\int_y^z \frac{\partial f}{\partial x}(x,t)dt-\int_{y_0}^{z_0} \frac{\partial f}{\partial x}(x,t)dt \right |=\left |\int_{z_0}^z \frac{\partial f}{\partial x}(x,t)dt - \int_{y_0}^{y} \frac{\partial f}{\partial x}(x,t)dt\right |\\ \le M|z-z_0|+M|y-y_0|\le 2M \|(x,y,z)-(x_0,y_0,z_0) \|$$ Where $M$ is the least upper bound of $f$ on the compact rectangle $I' \times J'$ and $\| . \|$ is the max norm on $\mathbb R^3$. This proves that $\frac{\partial \phi}{\partial x}$ is continuous (even Lipschitz continuous) on $I' \times J' \times J'$.
Hence $\phi$ has continuous partial derivatives on $I' \times J' \times J'$, and the chain rule gives the desired result.