Let
- $d\in\mathbb N$ and $U\subseteq\mathbb R^d$ be open
- $\tau>0$ and $T_t$ be a $C^1$-diffeomorphism from $U$ onto an open subset of $\mathbb R^d$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_U$ (assume $[0,\tau)\ni t\mapsto T_t(x)$ is differentiable for all $x\in\mathbb R^d$)
- $v_t:T_t(U)\to\mathbb R^d$ with $$v_t\left(T_t(x)\right)=\frac{\rm d}{{\rm d}t}T_t(x)\;\;\;\text{for all }(t,x)\in[0,\tau)\times U\tag1$$
- $\Omega\subseteq U$ and $$\Omega_t:=T_t(U)\;\;\;\text{for }t\in[0,\tau)$$
- $E$ be a $\mathbb R$-Banach space and $u_t:\Omega_t\to E$ be differentiable
If $$[0,\tau)\to E\;,\;\;\;t\mapsto u_t\left(T_t(x)\right)\tag2$$ is differentiable at $0$ for all $x\in\Omega$, then $${\rm d}u[v](x):=\left.\frac{\rm d}{{\rm d}t}u_t\left(T_t(x)\right)\right|_{t=0}\;\;\;\text{for }x\in\Omega$$ is called material derivative of $u$. On the other hand, if $$x\in\bigcap_{t\in[0,\:\delta)}\Omega_t\tag3$$ for some $\delta\in(0,\tau]$ and $$[0,\delta)\to E\;,\;\;\;t\mapsto u_t(x)\tag4$$ is differentiable at $0$, then $$u'[v](x):=\left.\frac{\rm d}{{\rm d}t}u_t\left(x\right)\right|_{t=0}$$ is called local shape derivative of $u$ at $x$.
If $u$ has a material derivative and a local shape derivative at $x\in\bigcap_{t\in[0,\:\delta)}\Omega_t$ for some $\delta\in(0,\tau]$, are we able to apply the chain rule and conclude $${\rm d}u[v](x)=u'[v](x)+\left\langle\nabla u_0(x),v_0(x)\right\rangle\tag5?$$
Note that, by $(1)$, $$v_0(x)=\frac{\rm d}{{\rm d}t}T_t(x)\tag6.$$