If there is a function $f(x,y)$ and $x=x(t),y=y(t)$ such that $f:R^2\to R$. I want to prove that $\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$ (All functions are differentiable).
Most of the proofs I've seen, use ideas like total differential (which I don't understand), or have steps which are not explained because of complexity. So I've been trying to prove it using single variable calculus ideas, and the partial derivative.
This is what I've got (mostly from an MIT_OCW video):
$\Delta f=\Delta f_x+\Delta f_y$ (Here $\Delta f$ is the change in $f$, when there is a change in $x$)
$\Rightarrow\Delta f=\frac{\Delta f_x}{\Delta x}\Delta x+\frac{\Delta f_y}{\Delta y}\Delta y$
$\frac{\Delta f_x}{\Delta x}\approx\frac{\partial f}{\partial x}$ and $\frac{\Delta f_y}{\Delta y}\approx\frac{\partial f}{\partial y}$
$\Rightarrow \Delta f\approx f_x\Delta x+ f_y\Delta y$
$\tag{1} \Rightarrow \frac{\Delta f}{\Delta t}\approx f_x\frac{\Delta x}{\Delta t}+ f_y\frac{\Delta y}{\Delta t}$
As $\Delta t\to 0$, $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$, so when I apply limit it's supposed to become:
$\lim_{\Delta t\to 0}\frac{\Delta f}{\Delta t}=\lim_{\Delta t\to 0}f_x\frac{\Delta x}{\Delta t}+\lim_{\Delta t\to 0}f_y\frac{\Delta y}{\Delta t}$
$\tag{2} \Rightarrow \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$
I know that as $\Delta t\to 0$, $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$, $\frac{\Delta y}{\Delta t}\to \frac{dy}{dt}$ and $\frac{\Delta f}{\Delta t}\to \frac{df}{dt}$, but how do I know that $(f_x\frac{\Delta x}{\Delta t}+ f_y\frac{\Delta y}{\Delta t})$ converges to $\frac{df}{dt}$?
(1) is an approximation, and (2) is an equality. How do I know that (1) turns into an equality when the limit is applied?
UPDATE 1: Keeping the error terms in the proof:
$\Delta f=\Delta f_x+\Delta f_y$
From manipulating the Taylor's series I get
$\Delta f_x=\frac{\partial f}{\partial x}\Delta x+\sigma_1(\Delta x)$ and
$\Delta f_y=\frac{\partial f}{\partial y}\Delta y+\sigma_2(\Delta y)$
(Here $\sigma_1(\Delta x)$ is a function of $\Delta x$, and $\to 0$ as $\Delta x\to 0$)
$\Rightarrow \Delta f=f_x\Delta x+f_y\Delta y+\sigma_1(\Delta x)+\sigma_2(\Delta y)$
$\Rightarrow \frac{\Delta f}{\Delta t}=f_x\frac{\Delta x}{\Delta t}+ f_y\frac{\Delta y}{\Delta t}+\frac{\sigma_1(\Delta x)}{\Delta t}+\frac{\sigma_2(\Delta y)}{\Delta t}$
As $\Delta t\to 0$, $\Delta x\to 0$ (If I consider $\Delta x=x(t+\Delta t)-x(t)$),
also $\frac{\Delta x}{\Delta t}\to \frac{dx}{dt}$, and $\frac{\sigma_1(\Delta x)}{\Delta t}\to 0$, as $\sigma_1(\Delta x)\to 0$ much faster than $\Delta t$
$\Rightarrow \lim_{\Delta t\to 0}\frac{\Delta f}{\Delta t}=\lim_{\Delta t\to 0}f_x\frac{\Delta x}{\Delta t}+\lim_{\Delta t\to 0}f_y\frac{\Delta y}{\Delta t}$
$\therefore \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$
What I'm unsure about are the Taylor series manipulations. This is what I did for $\Delta f_x$:
$f(x)=f(c)+f`(c)(x-c)+\sigma(x-c)$ (where $\sigma(x-c)$ = all higher order terms)
Taking $c=x$, and $x=x+\Delta x$. I get
$f(x+\Delta x)=f(x)+f`(x)(\Delta x)+\sigma(\Delta x)$
$\Rightarrow f(x+\Delta x)-f(x)=f`(x)(\Delta x)+\sigma(\Delta x)$
$\therefore \Delta f_x=f`(x)(\Delta x)+\sigma(\Delta x)$
I'd appreciate it if someone could check this proof and the validity of these manipulations.
