$0 = \infty{}{}{}{}{}$? I am just learning multivariable calculus and my teacher write this equasion on the board, something like
Given $F:\mathbb{R^2}\to\mathbb{R},\\x, y:\mathbb{R}\to\mathbb{R}\\t\in\mathbb{R}$
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and he said let x and y be arguments of F, and t is a parameter, and x and y are a paremetric equation of t, and then he wrote that
$$\frac{d}{dt}F(x,y) = \frac{dF}{dt} = \frac{dF}{dx}\cdot\frac{dx}{dt}+\frac{dF}{dy}\cdot\frac{dy}{dt}$$
and so i got really confused because in single variable calculus u can go
$\frac{dg}{dy}\cdot\frac{dy}{dx}$ and cross out the dy's because simple fraction multiplication and get $\frac{dg}{dx}$ and so here in the above equasion my instinct really wants to just cross out the dy's and dx's and get Equation One.
$$\frac{dF}{dt}=\frac{dF}{dt}+\frac{dF}{dt}$$
and so heres where the trouble starts. I can turn it into $$\frac{dF}{dt}=2\cdot\frac{dF}{dt}$$
$$\frac{dF}{dt}=0$$
but this is for ALL F(x,y) where x and y are function of t, and that doesnt seem right, but I can take it one step further and go $$\frac{dF}{dt} = \frac{dF}{dt}+\frac{dF}{dt}$$
and by recursive substitution, a sketchy but proven method for finding fixed points (or scalar solutions to recurrence relations) I can
$\frac{dF}{dt} = \frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}$
$\frac{dF}{dt} = \frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}$
$\frac{dF}{dt} = \frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}$
$\frac{dF}{dt} = \frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+\frac{dF}{dt}+⋯$
and the only solution for infinite things equal to eachother summed up is $$\frac{dF}{dt}=\infty$$ and so now we have $$0=\frac{dF}{dt}=\infty$$
$$0=\infty$$
but that doesn't seem right so I must have gone wrong somewhere. If anyone can help me, I am new to this and it would be much appreciated, thank you :)
 A: The total derivative gives
$$\dfrac{dF}{dt} = \dfrac{\partial F}{\partial x}\dfrac{dx}{d t}+\dfrac{\partial F}{\partial y}\dfrac{dy}{d t}$$
The partial derivative means something different than a "regular derivative". It means one variable is kept as the independent variable while all others are kept constant. This should help your intuition for why "cancellation" is not possible. Though, technically, while it works in nice cases, it isn't strictly algebraic, anyway.
A: You should check your notes.
Your teacher must have written something closer to $\frac{dF(x,y)}{dt} =\frac{\partial F(x,y)}{\partial x} \frac{dx}{dt} + \frac{\partial F(x, y)}{\partial y} \frac{dy}{dt}$
The partial differential notation is very important here. $F$ is a function of two independent variables, not just a single variable. The total change in $F$ is calculated by adding the individual contributions caused by the changes in each of the independent variables. That's basically what that equation says.
A: $\frac{dF}{dx}$should be $\frac{\partial F}{\partial x}$and $\frac{dF}{dy}$should be $\frac{\partial F}{\partial y}$. Then you will have the chain rule, one of the fundamental theorems of mulilinear calculus. Think of a model $z=F(x,y)$ and interpret the partial derivatives $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$ as derivatives of cross-sectional curves.   
