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# 2nd order differential substitution theory help please watch

1. Hello everyone.

So I don't seem to be understanding what is happening when I differentiate and as such I am getting into some serious muddles

So let's say we have a function of x, V, and y=V*x

we have a 2nd order differential equation of y in terms of x ... all good so far, hopefully?

So what I do then is I then do dy/dx (so I can sub it into my equation)

dy/dx = V + x*dv/dx (by the product rule)

and this is where I can't seem to progress.

I know that d^2y/dx^2 = 2dv/dx + x*d^2v/dx^2 but I don't understand why that is true.

When I differentiate the dv/dx wrt x ... as far as I can see, dv/dx is an implicit function.

This is relevant to me because If I had g^2 ... and I differentiate it, I get 2g dg/dx ... (I know why that is true)

So I don't understand why the chain rule stops "working" or something? What am I missing?

Why is d/dx (dv/dx) not d^2v/dx^2 * dv/dx ?
2. (Original post by DrSebWilkes)
Hello everyone.

So I don't seem to be understanding what is happening when I differentiate and as such I am getting into some serious muddles

So let's say we have a function of x, V, and y=V*x

we have a differential equation of y in terms of x ... all good so far, hopefully?

So what I do then is I then do dy/dx (so I can sub it into my equation)

dy/dx = V + dv/dx (by the product rule)

and this is where I can't seem to progress.

I know that d^2y/dx^2 = 2dv/dx + d^2v/dx^2 but I don't understand why that is true.

When I differentiate the dv/dx wrt x ... as far as I can see, dv/dx is an implicit function.

This is relevant to me because If I had g^2 ... and I differentiate it, I get 2g dg/dx ... (I know why that is true)

So I don't understand why the chain rule stops "working" or something? What am I missing?

Why is d/dx (dv/dx) not d^2v/dx^2 * dv/dx ?
Bold is your problem, you've misapplied the product rule. I'll just tell you because it should be obvious when you see it anyway:

I think you're missing an x from the thing you "know" as well.
3. (Original post by Rinsed)
Bold is your problem, you've misapplied the product rule. I'll just tell you because it should be obvious when you see it anyway:

I think you're missing an x from the thing you "know" as well.
Oh cheers. That was merely information I meant to add, sorry.

Still doesn't clear up the problem about how to differentiate the dv/dx term though ...
4. (Original post by DrSebWilkes)
Oh cheers. That was merely information I meant to add, sorry.

Still doesn't clear up the problem about how to differentiate the dv/dx term though ...
Apologies, I misunderstood quite what you were asking. Addressing the following question:

Why is d/dx (dv/dx) not d^2v/dx^2 * dv/dx ?

Because this is not an example of the product rule.

With your g^2 example, you can think about it two different ways. First, by the product rule:

Secondly, you can work directly from the chain rule. The answers are the same because mathematics requires it.

But with your example, d/dx(dv/dx), this is just differentiating V twice. This is the trivial case of the chain rule, you can just differentiate to get . We don't require a second part because there is no product involved.

Also think, by analogy with d/dx g^2, above: would give you the answer you say you expect, and they can't both be right.

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