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    Hi,

    The chain rule is: dy/dx = dy/dv * dv/dx
    So why in this question does it differentiate dv/dx to be d2v/dx2 . Where does the dy/dv go? Picture is attached.
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    (Original post by Seytonic)
    Hi,

    The chain rule is: dy/dx = dy/dv * dv/dx
    So why in this question does it differentiate dv/dx to be d2v/dx2 . Where does the dy/dv go? Picture is attached.
    Name:  IMG_20160603_084402.jpg
Views: 73
Size:  315.7 KB
    I'm not quite sure how you're applying the chain rule there.

    We have dy/dx = x(dv/dx) + v
    Then d^2y/dx^2 is just the derivative of that with respect to x.

    By the chain rule, the derivative of x(dv/dx) with respect to x is 1*(dv/dx) + x*(d^2v/dx^2) and of course the derivative of v with respect to x is just dv/dx. And you add them for the mark scheme answer
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    (Original post by Seytonic)
    Hi,

    The chain rule is: dy/dx = dy/dv * dv/dx
    So why in this question does it differentiate dv/dx to be d2v/dx2 . Where does the dy/dv go? Picture is attached.
    Name:  IMG_20160603_084402.jpg
Views: 73
Size:  315.7 KB
    You don't need the chain rule here.

    \displaystyle \frac{dy}{dx}=\frac{dv}{dx}x + v

    Differentiate both sides with respect to x using the product rule for the first term.

    And use the fact that

    \displaystyle \frac{d}{dx}\left(\frac{dv}{dx} \right ) = \frac{d^2 v}{dx^2}
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    (Original post by 13 1 20 8 42)
    I'm not quite sure how you're applying the chain rule there.

    We have dy/dx = x(dv/dx) + v
    Then d^2y/dx^2 is just the derivative of that with respect to x.

    By the chain rule, the derivative of x(dv/dx) with respect to x is 1*(dv/dx) + x*(d^2v/dx^2) and of course the derivative of v with respect to x is just dv/dx. And you add them for the mark scheme answer
    (Original post by notnek)
    You don't need the chain rule here.

    \displaystyle \frac{dy}{dx}=\frac{dv}{dx}x + v

    Differentiate both sides with respect to x using the product rule for the first term.

    And use the fact that

    \displaystyle \frac{d}{dx}\left(\frac{dv}{dx} \right ) = \frac{d^2 v}{dx^2}
    Thanks for your responses, I think I see where you're coming from, but why in this other question (which I got right) did my approach work? Pic is attached.
    EDIT: For the 2nd blue circle I meant to highlight dt/tx

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    (Original post by Seytonic)
    Thanks for your responses, I think I see where you're coming from, but why in this other question (which I got right) did my approach work? Pic is attached.
    Name:  IMG_20160603_091149.jpg
Views: 49
Size:  225.6 KB
    Sorry I meant product rule before. That approach is right because in that case chain rule is needed
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    (Original post by Seytonic)
    Thanks for your responses, I think I see where you're coming from, but why in this other question (which I got right) did my approach work? Pic is attached.
    EDIT: For the 2nd blue circle I meant to highlight dt/tx

    Name:  IMG_20160603_091149.jpg
Views: 49
Size:  225.6 KB
    In this case you have a derivative with respect to t on the right-hand-side so if you want to differentiate the whole thing with respect to x then you need the chain rule.

    In the previous question every derivative was respect to x so you can differentiate without the chain rule.
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    (Original post by notnek)
    In this case you have a derivative with respect to t on the right-hand-side so if you want to differentiate the whole thing with respect to x then you need the chain rule.

    In the previous question every derivative was respect to x so you can differentiate without the chain rule.
    Ahhh, that clears it up, cheers.
 
 
 
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