For real number functions (if they are differentiable) $f$ and $d$ it holds: $$\frac{\mathbf{d}}{\mathbf{d}t}d(f(t)) = d'(f(t))f'(t).$$
Now let $t$ be a real number but $v$, $f(t)$, and $d(v)$ be vectors. The above formula may not make sense because $d$ is now a function from a vector.
Please help to "save" the formula, that is define how it can be demonstrated a similar formula for vectors.
Maybe it may be described with something like gradient?
Let $t \in \mathbb{R}$, $f: \mathbb{R} \rightarrow X$, and $d: X \rightarrow Y$, where $X,Y$ are real Banach spaces.
$f$ is said to be differentiable at $t$ if there exists $f'(t) \in X$ and $E: \mathbb{R} \rightarrow X$ such that
$$ f(t + h) = f(t) + f'(t) h + \|h\| E(h)$$
for every $h$ in some open neighborhood of the origin of $\mathbb{R}$, and such that $\lim_{h \rightarrow 0} E(h) = 0$.
Similarly $d$ is said to be (Frechet) differentiable at $x \in X$ if there exists a bounded linear operator $d'(x): X \rightarrow Y$ and $E: X \rightarrow Y$ such that
$$ d(x + h) = d(x) + [d'(x)](h) + \|h\| E(h) $$
for every $h$ in some open neighborhood of the origin of $X$, and such that $\lim_{h \rightarrow 0} E(h) = 0$. Notice that the linear term $[d'(x)](h)$ is now the action of a linear operator $d'(x)$ on a vector $h$.
The chain rule states that if both $d,f$ are differentiable, then $d \circ f$ is differentiable and its derivative at $t$ is a linear map $(d \circ f)'(t): \mathbb{R} \rightarrow Y$ satisfying
$$ [(d \circ f)'(t)](h) = [d'(f(t))](f'(t) h)$$
for every $h$ in the neighborhood of the origin of $\mathbb{R}$.
In the scalar case, we don't need to add the "$h$" into the chain rule because all the derivatives just multiply against $h$ as a scalar. But in the full vector case, it is important to identify how the $h$ is actually placed into the derivative equation as an input.