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I am just learning multivariable calculus and my teacher write this equasion on the board, something like

Given $F:\mathbb{R^2}\to\mathbb{R},\\x, y:\mathbb{R}\to\mathbb{R}\\t\in\mathbb{R}$

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 :)

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    $\begingroup$ Yeah, this is a situation when trusting your gut isn't a great idea. In general you won't get $$\frac{dF}{dt}=\frac{\partial F}{\partial x}\cdot \frac{dx}{dt}$$ because $dF/dt$ takes changes in both $x$ and $y$ into account. The contributions of changes in $x$ and changes of $y$ to the total need to be added together to get the total change. $\endgroup$ – subrosar Feb 8 '20 at 3:00
  • $\begingroup$ Consider $F(x,y) = xy$, $x(t) = t$, and $y(t) = t$. Keep in mind that since $F(x,y)$ is a function of $x$ and $y$ you will need to use partial derivatives when differentiating $F(x,y)$ with respect to $x$ or with respect to $y$. $\endgroup$ – SOULed_Outt Feb 8 '20 at 3:01
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    $\begingroup$ Regarding your comment about "crossing out" differentials by using "simple fraction multiplication", you would do well to note that derivatives are not fractions and cannot be treated as such in general. $\endgroup$ – Brian Feb 8 '20 at 3:01
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    $\begingroup$ Also, I think it would help to point out that having multiple solutions to an equation does not necessarily imply that the solutions must be equivalent. You've argued that a solution is $0$ or $\infty$. That does not imply that they are equivalent. $\endgroup$ – SOULed_Outt Feb 8 '20 at 3:07
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    $\begingroup$ thanks everyone, especially souled_out who told be why i can have z=z+z then z=0 and z=∞ without 0=∞ because 0∈z and ∞∈z, i guess I may need to look out for whether I have found The solution for some arbitrary equation, or a solution satisfying the conditions but which is not necessarily the only one. I will need to check for all real numbers, all complex numbers, ∞, -∞, ∞i, -∞i, and probably these nonsense quaternion "numbers" too $\endgroup$ – cmarangu Feb 8 '20 at 5:06
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$\frac{dF}{dx}$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.

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  • $\begingroup$ yes i knew the partial derivative of a continuous differentiable function of $\mathbb{R^2}$ with respect to x gives you the derivative of the single variable function with y as an argument, but I am still confused as to how I am able to go from x=2x to 0=∞ can you help? $\endgroup$ – cmarangu Feb 8 '20 at 3:17
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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.

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  • $\begingroup$ oh right that makes more sense, how could I possible take a single-variable derivative of a multi-variable function? I will blame it on my teacher's bad handwriting, thanks. I am still confused as to how I was able to go from z=2z to 0=∞ though $\endgroup$ – cmarangu Feb 8 '20 at 3:19
  • $\begingroup$ If you know $x=a(t), y=b(t)$ then you can substitute them in. Then finding $dF/dt$ is as normal. $\endgroup$ – David P Feb 8 '20 at 3:51
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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.

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  • $\begingroup$ Yes of course I could not differentiate a multi-variable function with respect to only one variable x and expect the output to be a constant, it (the 'derivative' with respect to x) would be a function with y as an argument! The notation helps visualize this as I cannot as easily say $dx=\partial x$, however I am still wondering how I was able to go from $z=z+z$ to $0=∞$ $\endgroup$ – cmarangu Feb 8 '20 at 3:22
  • $\begingroup$ It's because your understanding was wrong, and your inappropriate use of the total derivative notation on the RHS confused you further. That's how you got a something (non-zero)= twice of itself result, which is absurd and can lead to any wrong result ("paradox") you want, including zero = infinity. $\endgroup$ – Deepak Feb 8 '20 at 3:27
  • $\begingroup$ oh. i was anticipating something more along the lines of peoples responses to youtu.be/CaasbfdJdJg?t=792 but your probably right never mind. $\endgroup$ – cmarangu Feb 8 '20 at 3:38

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