On proving the total differential. I am following an open-course on multi variable calculus provided by MIT taught by Denis Auroux. The question I am about to ask is from this lecture.
In the lecture Denis Arnoux gives a sketch proof of the so called total differential. Assume that we have a function $f(x,y,z) : R^3 \rightarrow R^3$ the total differential is
$$\frac{df}{dt} = f_{x}\frac{dx}{dt} + f_{y}\frac{dy}{dt} + f_{z}\frac{dz}{dt} $$
where the notation $f_x$ is the partial derivative with respect the variable of interest, in this case x. (of course the concept can be generalized to $R^n$).
The sketch proof begins by taking the approximation
$$\Delta f \approx f_x \Delta x + f_y \Delta y + f_w \Delta w \implies \frac{\Delta f}{\Delta t} \approx \frac{f_x \Delta x + f_y \Delta y + f_w \Delta w}{\Delta t}$$
Now noticing that taking the limit as $\Delta t \rightarrow 0$ we obtain for every ratio of $\Delta$ the definition of the derivative ( $\lim_{\Delta t \rightarrow 0} \Delta f / \Delta t := df/ dt$ )
we obtain $\frac{df}{dt} = f_{x}\frac{dx}{dt} + f_{y}\frac{dy}{dt} + f_{z}\frac{dz}{dt}$ because "the approximation gets better and better".
My question is: could somebody provide a rigorous proof of this fact following the sketch? Is the only problem the approximation sign turning into the equal sign?
 A: Hint.
The total differential of $f$ is obtained when you consider the fonction $$F: t \to f(x(t),y(t),z(t))$$ which is the composition of $$\gamma : t \to (x(t),y(t),z(t))$$ with $f$. And you can find $F$ derivative by applying the chain rule.
A: Well, first I'll assume $f$ is a $\mathbb{R^3}\rightarrow \mathbb{R}$ function, as if it were a vector-valued function, you'd be able to do everything component-wise pretty much.
Secondarily, if $f$ is a function of $(x,y,z)$, then it makes no sense to talk about $df/dt$, so I assume what we have here, is a smooth curve $\gamma:\mathbb{R}\rightarrow\mathbb{R}^3$, whose range is in $f$'s domain, then we can take $$ \left.\frac{df\circ\gamma}{dt}\right|_t. $$
Okay, now your problem, which is basically proving chain rule is as follows:
The curve's derivative obeys the following: $$ \gamma(t+h)-\gamma(t)=\dot{\gamma}(t)h+\varepsilon(h), $$ where epsilon is a function that tends to zero at very fast, when $h$ also tends to zero.
Then $$ \left.\frac{df\circ\gamma}{dt}\right|_t=\lim_{h\rightarrow0}\frac{f(\gamma(t+h))-f(\gamma(t))}{h}=\lim_{h\rightarrow0}\frac{f(\gamma(t)+\dot{\gamma}(t)h)-f(\gamma(t))}{h}. $$ But if we rename $\gamma(t)=(x(t),y(t),z(t))=(x,y,z)=x$, and $\dot{\gamma}(t)=v=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)$, then the above is $$ \left.\frac{df\circ\gamma}{dt}\right|_t=\lim_{h\rightarrow0}\frac{f(x+hv)-f(x)}{h}, $$ which is the $d_vf$ directional derivative of $f$ in the direction of $v$, and for (Fréchet) differentiable $f$, $$ d_vf=\frac{\partial f}{\partial x}v_1+\frac{\partial f}{\partial y}v_2+\frac{\partial f}{\partial z}v_3=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}. $$
