Question

Leibniz notation product rule... @ganeshie8 just be expected to be tagged

Answer 1

\[\frac{ d^2y }{ dx^2 }\] I'm trying to get this to equal \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\] so I have \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] which I got from the chain rule but how do I use the product rule here X_X

Answer 2

@freckles

Answer 3

Wait so are we saying y(x) and x(t) is that true?

Answer 4

Well I'm trying to figure out eulers equation \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\] and it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2

Answer 5

so I can use my substitution

Answer 6

Oh ok this makes more sense now.

Answer 7

wait are you not just allowed to substitute in \[y=At^n\] and solve?

Answer 8

No I have to use x = lnt

Answer 9

and it's telling me to do all that stuff lol

Answer 10

Oh alright then

Answer 11

Just want to know how to use the product rule in leibniz notation dont really care about the problem

Answer 12

f'g+g'f lol but I can't believe I don't know how to do this

Answer 13

\[y(x(t))\]
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]
\[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \]

Try to take it from here.

Answer 14

While you do that, I'll do it without logs to see how much faster it is.

Answer 15

It's the first part, \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right)\frac{ dx }{ dt }\] there's no changeeee, I was expecting there to be d^2y/dx^2 or something

Answer 16

Ultimately they'll get the same answer and this method you're using is more general so it's not without its merits it's just this is a pretty common one to solve.

Answer 17

Here's only the derivative part of the first part (notice we'll have a squared dx/dt when we multiply this back in):
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \]

Answer 18

Ok should it not be \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

Answer 19

If this bothers you instead let:

\[\frac{dy}{dx} = u(x(t))\]

Then when you do the derivative you have:

\[\frac{d}{dt}(u) = \frac{du}{dx} \frac{dx}{dt}\]

Then you can look at the du/dx term on its own as the second derivative of y wrt x.

Answer 20

Explain your reasoning for this:

\[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

Answer 21

Nvm I'm dumb, it's \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right) = \frac{ dy }{ dt }\frac{ dt }{ dx }\frac{ d }{ dt } = \frac{ d^2y }{ dt^2 }\frac{ dt }{ dx }\] is this what you mean

Answer 22

and not dx^2

Answer 23

Nope, you're taking the derivatives in a strange order, and this hanging d/dt that's by itself is kind of meaningless here.

Answer 24

I just used the chain rule, the d/dt is what is confusing me

Answer 25

\[y(x(t))\]
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]
\[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \]

Looking only at the red part:
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \]

make the u substitution I described earlier to get through this part.

Answer 26

If there's anything about using the product and chain rule above that confuses you say something now before moving forward.

Answer 27

Nooo, it's not really the substitution part, I mean \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right)\] it was this in general, so you're just writing \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \frac{ dx }{ dt }\right)\] + this thing again, oh I'm overthinking this, I just realized how simple it actually is

Answer 28

\[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

Answer 29

omggggggg kai

Answer 30

Where did this come from?

Answer 31

Explain why you're trying to use this: \[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

Answer 32

That's my original q

Answer 33

Isn't this what you're trying to solve? \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\]

Answer 34

\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right) = \frac{ d }{ dx }(\frac{ dy }{ dt })t+\frac{ d }{ dx }\left( t \right)\frac{ dy }{ dt }\]

Answer 35

I was trying to find d^y/dt^2 in terms of d^2y/dx^2 ahaha

Answer 36

I think I got it, I sort of just thought about this \[\frac{ d }{ dx }(u*v)=...\] I was just overthinking

Answer 37

No stop saying that lol, you can't claim that until you've finished solving the problem xD

Right now I'm trying to get us on the path of writing \(\frac{d^2y}{dt^2}\) in terms of x with the chain rule, but I'm not sure where you're going wrong so I'm having difficulty helping you out.

Answer 38

Oh I thought we were doing it in terms of d^2y/dt^2 the whole time

Answer 39

I don't know what you're doing so I don't know

Answer 40

lmao

Answer 41

\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] Explain where you're getting this from, cause I don't know. That doesn't mean it's wrong it just means you need to say words, I can't read your mind.

Answer 42

All I was doing is finding \[\frac{ d^2y }{ dt^2 }\] in terms of \[\frac{ d^2y }{ dx^2 }\] so I would get \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\]

Answer 43

Hmmm explain how you're supposed to solve this: " it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2"

So I am not seeing this here, so for example, dy/dt in terms of dy/dx would look like this, right:

\[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]

Answer 44

\[\frac{ d^2y }{ dx^2 }\] and started using the chain rule, \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt }\frac{ dt }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] and then I was trying to apply chain rule to \[\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\]

Answer 45

I started with \[\frac{ d^2y }{ dx^2 }\]

Answer 46

product rule not chain rule* omg

Answer 47

So you're writing \[\frac{d^2y}{dx^2}\] in terms of \[\frac{d^2y}{dt^2}\] then? And you're plugging in \[\frac{dt}{dx}=t\] is that right?

Answer 48

Yes!

Answer 49

This is fine then I think, it's just weird cause instead of looking it as y(x(t)) you're looking at it as y(t(x)) which is kind of backwards but in this case works since \(x= \ln t\) which is surjective.

Answer 50

Haha, I thought that's how you were suppose to do it, I read it on pauls online notes or something, I really have no notes for this, so I was sort of coming up with stuff myself

Answer 51

I mean the point of getting this is so I can set up my equation as \[\frac{ d^2y }{ dx^2 }+(\alpha - 1)\frac{ dy }{ dx }+\beta y=0\]

Answer 52

Right of course, but doing it like this is awkward since time isn't generally thought of as a function of position. And it has no meaning in periodic motion.

Answer 53

true

Answer 54

I wasn't really thinking of it that way, haha just some math problem

Answer 55

And the question stated, let x = lnt and calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2 ahah

Answer 56

Here, I'll just walk through how I did it:

\[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]
\[\frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} \left( \frac{dx}{dt} \right)^2 + \frac{dy}{dt} \frac{d^2x}{dt^2}\]
I guess we can do it now or later, so why not now:

\[\frac{dx}{dt} = \frac{1}{t}\]
\[\frac{d^2x}{dt^2} = \frac{-1}{t^2}\]

And then plugging all this in you get the same simplified differential equation since the \(t^2\) will disappear.

Answer 57

I messed up one piece but, oh well should have been a dy/dx on that last term I just noticed

Answer 58

At least this way is more straight forward to plug in (well in my opinion). However I wouldn't use this method at all lol. Best way is to notice you have a power next to the derivative which is basically compensating for a lost power by the power rule every derivative. Still pretty easy to spot.

\[t^2 y''+\alpha t y' + \beta y = 0\]

Now you can just pick \[y=At^n\] and plug it in:

\[n(n-1)+\alpha n + \beta = 0\]

Now you have a quadratic in n to solve for the two solutions, done!

Answer 59

Yo that's so simple

Answer 60

Yeah I already hate this method I have to use

Answer 61

Haha use the hate to fuel you through the tedious calculations and before you know it you'll be really fast. I feel like most of this discussion didn't have to happen I just had no idea what you were doing so we wasted a lot of time.

Answer 62

Yeah haha, well at least I know what I'm doing now X_X

Answer 63

I just realized how simple this was, I actually just misread something earlier, so was sort of confused haha xD