Let $x$ be a map, $x:\mathbb{R}\rightarrow\mathbb{R}^n$.
Let $V$ be a map, $V:\mathbb{R}^n\rightarrow\mathbb{R}$.
Then the derivative $$\dfrac{d}{dt}V(x(t))=\nabla V(x(t))\cdot x'(t)=\langle\ \nabla V(x(t)),\ x'(t)\ \rangle$$ is the above inner product. However, I am stuck on understanding the first equality. The analogy to the "standard case" where $\dfrac{d}{dx}f(g(x))=f'(g(x))g'(x),f:\mathbb{R}\rightarrow\mathbb{R},g:\mathbb{R}\rightarrow\mathbb{R}$ is clear, but I would appreciate a slightly more rigorous derivation where the dot product becomes more clear.
Edit: I'm thinking it would be most understandable to me if we could use something like $\dfrac{d}{dt}=\dfrac{d}{d\text{something}}\dfrac{d\text{something}}{dt}$ to make the derivative of $V$ clearer.
Simple case first:
One way to "derive" the chain rule for $f:\mathbb{R}\rightarrow\mathbb{R},g:\mathbb{R}\rightarrow\mathbb{R}$ is:
$$\begin{align} \dfrac{d}{dx}f(g(x))&=\dfrac{dg(x)}{dx}\dfrac{d}{dg(x)}f(g(x))\\ &=\dfrac{dg(x)}{dx}\dfrac{df(g(x))}{dg(x)}\\ &=\dfrac{dg(x)}{dx} \dfrac{df(u)}{du}\biggr\rvert_{u=g(x)}\\ &=g'(x)f'(g(x)) \end{align}$$
Where I have tried to make it obvious that differentiating a function by its argument is then evaluated at the same argument.
So for this question:
$$\begin{align} \dfrac{d}{dt}V(x(t)) &\stackrel{(1)}{=} \bigg(\dfrac{dx_1(t)}{dt}\dfrac{\partial}{\partial x_1(t)}+\cdots+\dfrac{dx_n(t)}{dt}\dfrac{\partial}{\partial x_n(t)}\bigg)V(x(t))\\ &= \dfrac{dx(t)}{dt}\cdot \bigg(\dfrac{\partial}{\partial x_1(t)},\dots,\dfrac{\partial}{\partial x_n(t)}\bigg)V(x(t))\\ &\stackrel{(2)}{=} x'(t)\cdot \nabla V(x(t))\\ \end{align}$$
where in $(2)$ we analogously take the "standard" gradient of $V$, then evaluate at $x(t)$. This is because when we distribute $V$ into the vector of partial derivatives, we get $\dfrac{\partial V(x_1,...,x_n)}{\partial x_i}$, which is a derivative with respect to the $i$th variable then evaluated at $x(t)$.
And in $(1)$ we are rewriting the total derivative in terms of a sum of partial derivatives of all the variables relevant to $V(x)$, the function we wish to differentiate in a standard way (i.e. derivatives with respect to its arguments). That is,
$$\dfrac{d}{dt}=\dfrac{dx_1}{dt}\dfrac{\partial}{\partial x_1}+\cdots+\dfrac{dx_n}{dt}\dfrac{\partial}{\partial x_n}$$