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Query about transformations of second order differential equations (FP2) Watch

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    Hello

    I'm finding Chapter 5 of FP2 pretty nice. We just finished it in class today but there is one thing bugging me: the fact that I don't think I was taught the basics of differentiation fully.

    Take this example:



    When differentiating e^u dy/dx with respect to u in the first line, I understand we use the product rule to get the first term in the second line. But then how do we obtain d2y/dx2 * dx/du by differentiating dy/dx with respect to u?

    Similarly,



    I understand the use of the product rule here again to get dy/dz(-sinx) as the second term in the second line.

    But how does the first term of the second line come in? I'm just quite confused on what's being differentiated with respect to what.

    I would very much appreciate an explanation of these two differentials. Thanks a lot
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    (Original post by Student403)
    Hello

    I'm finding Chapter 5 of FP2 pretty nice. We just finished it in class today but there is one thing bugging me: the fact that I don't think I was taught the basics of differentiation fully.

    Take this example:



    When differentiating e^u dy/dx with respect to u in the first line, I understand we use the product rule to get the first term in the second line. But then how do we obtain d2y/dx2 * dx/du by differentiating dy/dx with respect to u?

    Similarly,



    I understand the use of the product rule here again to get dy/dz(-sinx) as the second term in the second line.

    But how does the first term of the second line come in? I'm just quite confused on what's being differentiated with respect to what.

    I would very much appreciate an explanation of these two differentials. Thanks a lot

    First one: Chain rule. When differentiating dy/dx wrt u, by the chain rule this is equivalent to differentiating wrt x and then multiplying by dx/du

    Second one: I'm not completely sure what all your variables are. Does dz/dz = cos x presumably?
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    (Original post by 16Characters....)
    First one: Chain rule. When differentiating dy/dx wrt u, by the chain rule this is equivalent to differentiating wrt x and then multiplying by dx/du

    Second one: I'm not completely sure what all your variables are. Does dz/dz = cos x presumably?
    Thanks for the help on the first one

    Sorry I should have said:

    z = sinx

    so dz/dx = cosx

    Or dy/dx = cosx * dy/dz
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    standard chain rule

    d/dx(dy/dt) = d2y/dt2 . dt/dx

    d/dx(dv/dw) = d2v/dw2 . dw/dx

    d/dv(dy/dx) = d2y/dx2 . dx/dv
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    (Original post by Student403)
    Thanks for the help on the first one

    Sorry I should have said:

    z = sinx

    so dz/dx = cosx

    Or dy/dx = cosx * dy/dz
    No problem. The principle is the same for the second one then. Using what we just discussed, how do you differentiate \frac{dy}{dz} wrt x?
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    (Original post by 16Characters....)
    No problem. The principle is the same for the second one then. Using what we just discussed, how do you differentiate \frac{dy}{dz} wrt x?
    Would that then be d2y/dz2 * dz/dx?


    (Original post by TeeEm)
    standard chain rule

    d/dx(dy/dt) = d2y/dt2 . dt/dx

    d/dx(dv/dw) = d2v/dw2 . dw/dx

    d/dv(dy/dx) = d2y/dx2 . dx/dv
    Thanks sir - those different examples clear it up very nicely
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    (Original post by Student403)
    Would that then be d2y/dz2 * dz/dx?



    Thanks sir - those different examples clear it up very nicely
    if not ...
    http://www.thestudentroom.co.uk/show....php?t=3875341
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    :rofl:

    I'd much prefer madasmaths to mad maths
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    (Original post by Student403)
    Thanks for the help on the first one

    Sorry I should have said:

    z = sinx

    so dz/dx = cosx

    Or dy/dx = cosx * dy/dz
    \displaystyle \frac{dy}{dx} = \cos x \frac{dy}{dz}

    Differentiate both sides w.r.t x

    \displaystyle \frac{d^2y}{dx^2} = \cos x \times \frac{d}{dx} \left(\frac{dy}{dz}\right) + \frac{d}{dx}(\cos x) \times \frac{dy}{dz}

    But \displaystyle \frac{d}{dx} \frac{dy}{dz} = \left(\frac{d}{dz} \frac{dy}{dz}\right) \times ?
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    (Original post by Student403)
    Would that then be d2y/dz2 * dz/dx?
    Yep, and you know that dz/dx = cos x.

    Also, damn, got ninjad like 5 times over. :rofl:
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    (Original post by Zacken)
    \displaystyle \frac{dy}{dx} = \cos x \frac{dy}{dz}

    Differentiate both sides w.r.t x

    \displaystyle \frac{d^2y}{dx^2} = \cos x \times \frac{d}{dx} \left(\frac{dy}{dz}\right) + \frac{d}{dx}(\cos x) \times \frac{dy}{dz}

    But \displaystyle \frac{d}{dx} \frac{dy}{dz} = \left(\frac{d}{dz} \frac{dy}{dz}\right) \times ?
    (Original post by Zacken)
    Yep, and you know that dz/dx = cos x.

    Also, damn, got ninjad like 5 times over. :rofl:
    Nah I find your LaTeX very clear and helpful I think I finally get it!

    I reckon one of the main reasons I had this problem was because I wasn't taught exactly what the difference between things like "d/dx" and "dy/dx" were until very recently, so it hasn't fully fit itself in to my understanding of basic calculus


    Thank you all so much for the help
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    (Original post by Student403)
    Would that then be d2y/dz2 * dz/dx?
    Correct, so when this is subbed into your product rule you will get what you want.

    Edit: Whoops very slow
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    (Original post by 16Characters....)
    Correct, so when this is subbed into your product rule you will get what you want.

    Edit: Whoops very slow
    Much appreciated
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    (Original post by Student403)
    Nah I find your LaTeX very clear and helpful I think I finally get it!

    I reckon one of the main reasons I had this problem was because I wasn't taught exactly what the difference between things like "d/dx" and "dy/dx" were until very recently, so it hasn't fully fit itself in to my understanding of basic calculus
    Confession time: I've only learnt this myself very recently, like a few weeks ago. I'm not sure how I got by without knowing it before that :lol:
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    (Original post by Zacken)
    Confession time: I've only learnt this myself very recently, like a few weeks ago. I'm not sure how I got by without knowing it before that :lol:
    It's a real shame, isn't it? differentiation was a topic in GCSE maths fgs >.> Least they could have done was bring up the basic principles in this much detail in C1/C2
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    You can think of it as  \displaystyle \frac{d}{dx}(\frac{dy}{dz})= \frac{dz}{dx}  \frac{d}{dz}(\frac{dy}{dz})

    which I think should make it easier to see what's going on.
    It's almost as if the dx's cancel out leaving the differentiating the function with respect to z and multiplying through by dz/dx.
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    (Original post by B_9710)
    You can think of it as  \displaystyle \frac{d}{dx}(\frac{dy}{dz})= \frac{dz}{dx}  \frac{d}{dz}(\frac{dy}{dz})

    which I think should make it easier to see what's going on.
    It's almost as if the dx's cancel out leaving the differentiating the function with respect to z and multiplying through by dz/dx.
    Not just almost but exactly. dy/dz and dz/dx are quotients of infinitesimals, and by the Transfer Principle, they obey all the same rules as real numbers, so can be cancelled as normal.
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    (Original post by constellarknight)
    Not just almost but exactly. dy/dz and dz/dx are quotients of infinitesimals, and by the Transfer Principle, they obey all the same rules as real numbers, so can be cancelled as normal.
    That's not how we think of dy/dx and dx/dx anymore. Thinking of them as quotients of infinitesimals is old-school.
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    (Original post by Zacken)
    That's not how we think of dy/dx and dx/dx anymore. Thinking of them as quotients of infinitesimals is old-school.
    Non-standard analysis was invented in the 1960s - not that long ago!
 
 
 
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