Differentiating an integral with moving endpoints

Localisation. Fix $t_0\in I$. Choose a compact interval $K\subset I$ with $t_0$ in its interior $I_0$. The endpoint functions are bounded on $K$ by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, so choose $a<b$ in $J$ with $a<\alpha (t)<\beta (t)<b$ for $t\in K$, and choose $r_{\ast }\in J$ with $r_{\ast }<a$. Define $\mathrm{\Phi }(u,t):={\int }_{r_{\ast }}^{u}F(t,y)dy$ on $(a,b)\times I_0$. By [F1], $G(t)=\mathrm{\Phi }(\beta (t),t)-\mathrm{\Phi }(\alpha (t),t)$.

The primitive is $C^1$ on an open neighbourhood of the endpoint curves. By [F1], ${\mathrm{\partial }}_u\mathrm{\Phi }(u,t)=F(t,u)$; by [F2] on compact rectangles inside $I\times J$, ${\mathrm{\partial }}_t\mathrm{\Phi }(u,t)={\int }_{r_{\ast }}^u{\mathrm{\partial }}_{t}F(t,y)dy$. This last expression is jointly continuous: on a fixed compact rectangle, uniform continuity of ${\mathrm{\partial }}_{t}F$ (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous) bounds the change in $t$ by the interval length times a uniform error, and boundedness bounds the change in $u$ by a constant times $\mathrm{\mid }u-u_0\mathrm{\mid }$. Thus both partial derivatives are continuous, and If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative makes $\mathrm{\Phi }$ totally differentiable.

Apply [F3] to the curves $t\mapsto (\beta (t),t)$ and $t\mapsto (\alpha (t),t)$ in the open domain of $\mathrm{\Phi }$, and subtract using [F4]. Evaluation of the primitive gives $G^{\mathrm{\prime }}(t)=F(t,\beta (t)){\beta }^{\mathrm{\prime }}(t)-F(t,\alpha (t)){\alpha }^{\mathrm{\prime }}(t)+{\int }_{\alpha (t)}^{\beta (t)}{\mathrm{\partial }}_{t}F(t,y)dy$. The same uniform estimate as in step 1.2 shows this derivative is continuous. Since $t_0$ was arbitrary, the formula holds throughout $I$.