Equivalence of the two directional derivative definitions, without multivariable chain rule I am currently trying to nail down a good proof of the multivariable chain rule. So far, I have that for a function $ f: \mathbb{R}^n \to \mathbb{R}$ and a function $v: \mathbb{R} \to \mathbb{R}^n$, $$\dfrac{d}{dt} (f \circ v)(t) = \lim_{h \to 0} \dfrac{f(v(t)+hv'(t))-f(v(t))}{h}$$ This is, by definition, the directional derivative $\nabla_{v'(t)} f(v(t))$, which using the second definition of the directional derivative, is nothing but $\nabla f(v(t)) \cdot v'(t)$, which yields the multivariable chain rule. However, I cannot find a proof of the equivalence of these two definitions which does not use the multivariable chain rule in some way; so I ask:
1) Does a proof of this equivalence exist without the multivariable chain rule, or
2) If this equivalence cannot be shown without the multivariable chain rule, are there any nice proofs of which you know (I have yet to find one I like, besides this one if it does work out).  
 A: The equivalence is only true when $f$ is differentiable at that point $\vec v (t)$. If it is differentiable, then by definition of differentiability there exists a linear map $F: \Bbb{R}^n \to \Bbb{R}$ such that:
$$
\lim_{h \to 0} || f(\vec x + h\vec u) - f(\vec x) - F(h\vec u) || / h = 0
$$
Where $\vec u$ is a unit vector. This $F$ is precisely what we define as the gradient $\nabla f $ (actually, it's a row vector which is the transpose of the gradient vector). You can indeed show that yourself by taking the directional derivative at each of the basis vectors and checking that every component equals the $i$-th partial derivative.
If that is done, the proof goes through nicely supposing that the function $v$ is also differentiable where necessary, however that amounts to its derivative with respect to time exist at the point of interest, $t$. This means that every component must have a derivative at $t$.
Here is the proof of the above:
$$
\lim_{h \to 0} || f(\vec x + h\mathbf e_1) - f(\vec x) - F(h\mathbf e_1) || / h = || \frac{\partial f}{\partial x_1}(\vec x) - F(\mathbf e_1) || = 0 \\
\implies \frac{\partial f}{\partial x_1}(\vec x) = F(\mathbf e_1)
$$
Since this is a linear function $F: \Bbb{R}^n \to \Bbb{R}$, it's immediately seen to be a row vector.
$$
F(p) = \vec F \cdot \vec p
$$
Moreover, the right hand side is just the first component of this vector ($F(\mathbf e_1) = F_1$). Analogously we can do for all the other components.
EDIT: I will go into more details here but no, I'm afraid you can't get out of it without chain rule for at least single variable.
By differentiability of $v$ we can write that:
$$
\lim_{h \to 0} v(t + h) - v(t) = hv'(t)
$$
Thus we can write that:
$$
\lim_{h \to 0} || f(v(t+h)) - f(v(t)) - F(h\vec u) || / h = \\ \lim_{h \to 0} || f(v(t) + hv'(t)) - f(v(t)) - F(hv'(t)) || / h = 0
$$
Substitution of the limit there is guaranteed by continuity of $v$ at the point, which then equals your form.
