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?
Best Answer
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.