I would like to ask something related to the application of the differentiation under the integral sign (Leibniz rule) given by $$\dfrac{d}{dt}\int_{g(t)}^{h(t)}F(x,t)~dx=\left\{\dfrac{dh(t)}{dt}F(h(t),t)-\dfrac{dg(t)}{dt}F(g(t),t)\right\}+\int_{g(t)}^{h(t)}\dfrac{\partial}{\partial t}F(x,t)~dx\qquad(1)$$ Consider the signed power function $\mbox{sig}(x)^{\alpha}:=|x|^{\alpha}\mbox{sgn}(x)$ where $|x|$ is the absolute value of $x$ and $\mbox{sgn}(x)$ is the standard signum function. If the function to be integrated, i.e. $F(x,t)$ is given by $$F(x,t)=\mbox{sig}\left(\mbox{sig}\left(x\right)^{p}-\mbox{sig}\left(g(t)\right)^{p}\right)^{2-\frac{1}{p}}\qquad\qquad(2)$$ where $p\in\mathbb{R}$ is such that $1<p<2$ and the upper limiting function, $h(t)$ is smooth and the lower limiting function, $g(t)$ of the integral is given by $$g(t)=-\mbox{sig}\left(\omega(t)\right)^{\frac{1}{p}}\qquad\qquad(3)$$ with $p$ as given before and $\omega(t)$ is $C^{1}\left(\mathbb{R},\mathbb{R}\right)$, i.e. it is continuous but only one time differentiable. When the Leibniz rule is applied we obtain, $$\dfrac{d}{dt}\int_{g(t)}^{h(t)}F(x,t)~dx=\left\{\dot{h}(t)\mbox{sig}\left(\mbox{sig}\left(h(t)\right)^{p}-\mbox{sig}\left(g(t)\right)^{p}\right)^{2-\frac{1}{p}}+\frac{1}{p}|\omega(t)|^{\frac{1}{p}-1}\dot{\omega}(t)\mbox{sig}\left(\mbox{sig}\left(g(t)\right)^{p}-\mbox{sig}\left(g(t)\right)^{p}\right)^{2-\frac{1}{p}}\right\}+\left(2-\frac{1}{p}\right)\dot{\omega}(t)\int_{g(t)}^{h(t)}|\mbox{sig}\left(x\right)^{p}+\omega(t)|^{1-\frac{1}{p}}~dx\qquad\qquad(4)$$ where the second term in the curly brackets is problematic when $\omega(t)$ becomes equal to zero for some $t$, resulting in the indefinite form, $0/0$. To the best of my knowledge, in order to be able to apply this rule, both $h(t)$ and $g(t)$ should be continuously differentiable, whereas in my case $g(t)$ is only continuous. Does this mean that Leibniz rule is not applicable for the function given by (2) or is there another way to show that the integral $$\dfrac{d}{dt}\int_{g(t)}^{h(t)}F(x,t)~dx$$ with the function (2) and upper and lower limits mentioned before can still be computed?
Thanks a lot and best regards.