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2nd order differential substitution theory help please watch

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    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 ?
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    (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: \frac{dy}{dx} = V + x \frac{dV}{dx}

    I think you're missing an x from the thing you "know" as well.
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    (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: \frac{dy}{dx} = V + x \frac{dV}{dx}

    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 ...
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    (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:

     \frac{d}{dx} g(x)g(x) = g(x) \frac{dg}{dx} + \frac{dg}{dx} g(x) = 2 g(x) \frac{dg}{dx}

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

     \frac{d}{dx} g(x)^2 = 2 g(x) \frac{dg}{dx}

    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 \frac{d^2 v}{dx^2}. We don't require a second part because there is no product involved.

    Also think, by analogy with d/dx g^2, above:  \frac{d}{dx} (\frac{1}{2} (\frac{dv}{dx})^2) would give you the answer you say you expect, and they can't both be right.
 
 
 
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