Question

How could I intuitively figure out that \[\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{dx}{dt}+\frac{\partial f}{\partial y} \frac{dy}{dt}\] I feel like the linear independence of the variables is what makes this work so nicely but I'm not sure I could fully just figure this out on the fly if I wanted to.

Answer 1

I can try and help you

Answer 2

Sure sounds good

Answer 3

One moment while I generate an solution.

Answer 4

I'm trying to do a transformation from x and y to r and theta to see what the derivative would be to see if I get any kind of insight out of it.

Answer 5

Give me 5-7 more minutes

Answer 6

When T is dependent on x and y i.e t(x,y) then on the differentiation of t is done using chain rule

Answer 7

take x=rcos theta and y=rsin theta

Answer 8

Can you tell me exactly what you need me to find?

Answer 9

Well I just did it but I mean I'm really just blithely following the rule I've been given. It's not really giving me any insight into why this is the derivative of a multivariable function. I mean without checking, it could just as easily be something like \[\frac{df}{dt}=\frac{\partial ^2 f}{\partial x \partial y}* \frac{dx}{dt}* \frac{dy}{dt} \] You see what I mean?

Answer 10

Oh yeah I see

Answer 11

Do you want to try and solve it?

Answer 12

Yeah, I'm doing that right now actually lol.

Answer 13

Okay :) I'm going to be trying as well.

Answer 14

I'm sort of playing around and seeing what this differential equation can do for me \[f_{xy}x'y'=f_x x'+f_y y'\] so the left side is my proposed definition of what it should be and the right side is what it is already.

For clarity, f(x,y) with x(t) and y(t). So they're independent variables. No y(x) thing here.

Answer 15

Ohhh, I see what you're doing. I'll show you mines in one moment

Answer 16

I think it is kinda a hard nonlinear differential equation to solve with no general solution so... Although interesting it doesn't really get me anywhere lol. I don't know if I really know what you're doing.

Answer 17

\[\frac{ 2t }{ 2t }=\frac{ 2t }{ 2x } \frac{ dn }{ dt }+\frac{ 2t }{ 2y }\frac{ dy }{ dt }\]

Here t is an function of x & y \[\frac{ 2t }{ 2x}\]is the partial derivation occurs only for x and rest of the variables are constant
But \[\frac{ dt}{ dx } \] is the implicit derivative which occurs for both variables x & y with respect to x.

Answer 18

There is much difference between partial and implicit derivation

Answer 19

I'm kinda lost with the 2's and the n. I'm assuming that n should be an x and the 2s are supposed to be partials.

Answer 20

That's an example.

Answer 21

You've lost me.

Answer 22

Oh no!!

Answer 23

I gotta go for now, I'll be back either later tonight or tomorrow! =)

Answer 24

Alright :) Have a nice evening