Equivalence of functional and partial derivatives

I am trying to derive Newton's second law from the principle of least action, that is, setting the functional derivative $\frac{\delta S}{\delta x(t)}$ equal to 0. $$S = \int dt' \left[ \frac{m}{2} \left( \frac{dx}{dt'} \right)^2 - V(x(t')) \right] \tag{1}$$ So, \begin{align} \frac{\delta S}{\delta x(t)} &= \int dt' \left[ \frac{m}{2} \frac{\delta}{\delta x(t)} \left( \frac{dx}{dt'} \right)^2 -\frac{\delta V(x(t'))}{\delta x(t)} \right] \tag{2} \\ &= \int dt' \left[ m \frac{dx}{dt'}\frac{d}{dt'}\delta(t-t') - \frac{\delta V(x(t'))}{\delta x(t')}\frac{\delta x(t')}{\delta x(t)} \right] \tag{3} \\ &= - \int dt' \left[ m \frac{d^2x}{dt'^2}\delta(t-t') + \frac{\delta V(x(t'))}{\delta x(t')} \delta (t-t') \right] \tag{4} \\ &= -\left[ m \frac{d^2x}{dt^2} + \color{Red}{\frac{\delta V(x(t))}{\delta x(t)}} \right]. \tag{5} \\ \end{align}

Now that I have calculated $(5)$, and then set the variation of the action equal to zero, I know that $\frac{\delta V(x(t))}{\delta x(t)}$ must be the same as $\frac{\partial V(x(t))}{\partial(x(t))}$ in order to reproduce Newton's second law. How does the functional derivative turn into the partial derivative in this case?

Note: to get the second term in $(3)$, I used chain rule, but for functional derivatives.

The definition of the (integral of the) functional derivative (at least a definition that's good enough for physics level rigor) is the difference of the functional evaluated on a path $x(t)$ plus an arbitrary variation $\epsilon(t)$ and the functional evaluated on the path, to leading order in $\epsilon$. In other words \begin{equation} S[x(t)+\epsilon(t)]-S[x]=\int dt \frac{\delta S}{\delta x} \epsilon(t) + O(\epsilon^2) \end{equation} The fact that this definition puts the functional derivative inside of an integral is a reflection of the fact that the functional derivative is a distribution, like a Dirac delta function, it is only well defined inside of an integral.

Now define \begin{equation} S_V[x(t)]=\int dt V(x(t)) \end{equation} Then \begin{equation} S_V[x(t)+\epsilon(t)]=\int dt V(x+\epsilon)=\int dt\left( V(x) + \frac{\partial V}{\partial x}\epsilon+O(\epsilon^2)\right) \end{equation} Comparing with the definition of the functional derivative, we see we can identify \begin{equation} \frac{\delta S_V}{\delta x} = \frac{\partial V}{\partial x} \end{equation} which is the statement you need.

Here's how I think about it. An action is a functional: It eats a function and returns a number. The functional derivative asks: "For very small changes in the function fed to the functional, how how the functionals value change?"

First let's think of a trajectory, $x(t)$. This is what we will feed to the functional. Now let's consider a smooth family of such trajectories, $x_\lambda (t)$. That is, for each $\lambda$ we have a different trajectory, with small changes in $\lambda$ leading to small changes in $x_\lambda (t)$. Assume, in fact that there is a function $\delta x (t)$ such that

$$\delta x (t) = \lim_{\lambda \to 0} \frac{x_\lambda (t) - x_0 (t)}{\lambda}.$$

If each $\lambda$ gives a trajectory, each trajectory gives a real number when fed to a functional, then composition gives a function

$$S[x_\lambda]: \mathbb{R} \to \mathbb{R}$$.

This is just a real function, so we can take its derivative without any navel gazing.

If $S$ is nice enough, then there is a function which we will tantalizingly refer to as $\frac{\delta S}{\delta x}$ such that for any family $x_\lambda$, we have

$$\left.\frac{d S[x_\lambda]}{d \lambda}\right|_{\lambda = 0} = \int \frac{\delta S}{\delta x} \delta x \,dt.$$

So let's deal with a really simple "action" that is all potential:

$$S[x] = \int_{t_i}^{t_f} (V \circ x)(t) \, dt$$

I give you a function $x(t)$, you compose it with V, integrate it, and out pops a real number. If I give you a family of $x_\lambda$, then we have a function

$$S[x_\lambda] = \int_{t_i}^{t_f} (V \circ x_\lambda)(t) \, dt$$

Each $\lambda$ gives a different function, and therefore a different number. It's just a vanilla $\mathbb{R} \to \mathbb{R}$ function. Taking its derivative gives

$$\left.\frac{d S[x_\lambda]}{d \lambda}\right|_{\lambda = 0} = \frac{d }{d \lambda}\int_{t_i}^{t_f} (V \circ x_\lambda)(t) \, dt\\ = \int_{t_i}^{t_f} \frac{d }{d \lambda} (V \circ x_\lambda)(t) \, dt\\ = \int_{t_i}^{t_f} (V^\prime \circ x) \left( \left.\frac{d x_\lambda}{d \lambda}\right|_{\lambda = 0} \right) \, dt\\ = \int_{t_i}^{t_f} (V^\prime \circ x) \delta x \, dt$$

So that by looking at our definition we see that

$$\frac{\delta S}{\delta x} = V^\prime \circ x.$$

Note that the penultimate line follows just from chain rule.