Let $u \in C^\infty((0,T); H^1_0(\Omega))$ where $\Omega$ is bounded and smooth. Let $f\colon (0,T)\times \Omega \to \Omega$ be smooth with $f_t$ and $f$ uniformly bounded pointwise. Here $f_t$ refers to differentiation with respect to the first argument.
Is it true that $$\frac{d}{dt} (u(t)(f(t,x))) = u_t(t)(f(t,x))+ \nabla u(t)(f(t,x)) \cdot f_t(t ,x)$$
The $\nabla u$ means the weak gradient btw.
How could one prove this?
If $x$ is fixed let $b : (0,T)\times(0,T)\to \mathbb R$ defined by $$b(s, v) = u(s)(f(v, x))$$
It is clear that:
$$\frac{\partial b}{\partial s}(s, v) = u_t(s)(f(v,x))$$ and by the chain rule $$\frac{\partial b}{\partial v}(s, v) = \nabla u(s)(f(v,x)) \cdot f_t(v,x)$$ and $a(t) = u(t)(f(t,x)) = b(t, t)$ then by chain rule
\begin{align} \frac{\mathrm d}{\mathrm dt} a(t) &= \frac{\mathrm d}{\mathrm dt} b(t, t)\\ &= \frac{\partial b}{\partial s}(t, t)\frac{\partial s}{\partial t} + \frac{\partial b}{\partial v}(t, t)\frac{\partial v}{\partial t}\\ &= u_t(t)(f(t,x)) + \nabla u(t)(f(t,x))\cdot f_t(t,x) \end{align}