Consider $g$ continuous on $[a,b]\times \mathbb{R}^n$, $[a,b]\subset \mathbb{R}$, and $x\mapsto g(t,x)$ continuously differentiable for all $t\in [a,b]$.
I wonder if $t\mapsto \frac{\partial g}{\partial x}(t,x)$ is continuous on $[a,b]$.
Let $(t_n)$ a subsequence converging to $t\in [a,b]$. We have : $\frac{\partial g}{\partial x_i}(t_n,x) = \lim_{h\to 0}\frac{g(t_n,x+h e_i)-g(t_n,x)}{h}$.
So the question is can I have : $$ \lim_h \lim_n \frac{g(t_n,x+he_i)-g(t_n,x)}{h}= \lim_n\lim_h \frac{g(t_n,x+he_i)-g(t_n,x)}{h} $$
If the convergence $g(t,x+he_i)\to g(t,x)$ as $h\to 0$ is uniform then I have the above equality but even if $x\mapsto g(t,x)$ is continuous the limit may not be uniform (Dini's theorem). So it seems that there exists examples where $t\mapsto \frac{\partial g}{\partial x}(t,x)$ is not continuous on $[a,b]$.
edit : indeed the functions $g$ defined on $\mathbb{R} \times \mathbb{R}$ by $g(t,x) = tx^3\sin(\frac{1}{tx})$ is such an example.
The function $g$ defined on $\mathbb{R} \times \mathbb{R}$ by $g(t,x) = tx^3\sin(\frac{1}{tx})$ doesn't have continuous partial derivative at $t=0$