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

- Astrophysics

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

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- Astrophysics

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

- Astrophysics

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- Astrophysics

@freckles

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## More answers

- Kainui

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

- Astrophysics

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

- Astrophysics

so I can use my substitution

- Kainui

Oh ok this makes more sense now.

- Kainui

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

- Astrophysics

No I have to use x = lnt

- Astrophysics

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

- Kainui

Oh alright then

- Astrophysics

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

- Astrophysics

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

- Kainui

\[y(x(t))\]
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]
\[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \]
Try to take it from here.

- Kainui

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

- Astrophysics

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

- Kainui

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.

- Kainui

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

- Astrophysics

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

- Kainui

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.

- Kainui

Explain your reasoning for this:
\[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

- Astrophysics

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

- Astrophysics

and not dx^2

- Kainui

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

- Astrophysics

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

- Kainui

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

- Kainui

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

- Astrophysics

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

- Astrophysics

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

- Astrophysics

omggggggg kai

- Kainui

Where did this come from?

- Kainui

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

- Astrophysics

That's my original q

- Kainui

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

- Astrophysics

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

- Astrophysics

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

- Astrophysics

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

- Kainui

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.

- Astrophysics

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

- Kainui

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

- Astrophysics

lmao

- Kainui

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

- Astrophysics

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

- Kainui

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

- Astrophysics

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

- Astrophysics

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

- Astrophysics

product rule not chain rule* omg

- Kainui

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?

- Astrophysics

Yes!

- Kainui

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.

- Astrophysics

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

- Astrophysics

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

- Kainui

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.

- Astrophysics

true

- Astrophysics

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

- Astrophysics

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

- Kainui

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.

- Kainui

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

- Kainui

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!

- Astrophysics

Yo that's so simple

- Astrophysics

Yeah I already hate this method I have to use

- Kainui

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.

- Astrophysics

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

- Astrophysics

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

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