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Math Help - d(dy/dt)/dy

  1. #1
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    d(dy/dt)/dy

    Can someone please explain why \frac{\partial}{\partial y}(y'(x)), where y is a function of x, is necessarily zero? Is it because holding x in place forces y' to never change?
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    Re: d(dy/dt)/dy

    Hey phys251.

    Using the chain rule we get d/dy (y'(x)) = dy'(x)/dy * dy/dy = 0 * 1 = 0. I'm not sure if this answer suffices or not to answer your question, but the chain rule is always a good tool when you have these kinds of derivatives.
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    Re: d(dy/dt)/dy

    dy/dy is obvious, but I still don't get why dy'(x)/dy = d(dy/dx)/dy is zero.
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    Re: d(dy/dt)/dy

    Its a function of x so its like differentiating a constant.
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    Re: d(dy/dt)/dy

    Quote Originally Posted by chiro View Post
    Its a function of x so its like differentiating a constant.
    So, because the original equation was a partial, x remains constant, ergo y'(x) is constant too? If so, the partial with respect to y is definitely zero.
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    Re: d(dy/dt)/dy

    That is correct: since the expression doesn't involve y its like differentiating with respect to a constant term (which is zero).
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    Re: d(dy/dt)/dy

    Hi !

    Sorry to say : I don't agree with your line of reasoning
    Just look at the example in attachment and see that the result is not equal to zero.
    I let you think about for a while before giving the correct answer.
    Attached Thumbnails Attached Thumbnails d(dy/dt)/dy-example.jpg  
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    Re: d(dy/dt)/dy

    To finish what JJacquelin started: if y is not a function of x or related to x, then my post holds but yes y is related to x (so obvious once it is pointed out).

    You can use the inverse function theorem to make more specific claims.
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  9. #9
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    Re: d(dy/dt)/dy

    Quote Originally Posted by chiro View Post
    To finish what JJacquelin started: if y is not a function of x or related to x, then my post holds but yes y is related to x (so obvious once it is pointed out).
    Of course, if y is not a function of x or related to x then y'(x)=0 and there is no need for further discussion !
    The aim of my first post at arousing the controversy because the wording of the question was ambiguous.
    Introducing the partial derivative of (dy/dx) relatively to y suppose that dy/dx might be function of y, i.e.: dy/dx = f(x,y). There are many different manner to write f(x,y) and all are equal.
    For example, in the case y(x)=x
    f(x,y) = 2x, or f(x,y)=2sqrt(y), or f(x,y)=x+sqrt(y), or f(x,y)=2y/x, or etc.
    The partial derivative relatively of y is different for each one of these functions f(x,y). So, without more specification in the wording, the question is ambiguous.
    Among many possibilities, if we say : y(x) relates y to x as well as the inverse function x(y), then dy/dx is function of x which is function of y. Hence dy/dx is function of y. On this point of view, the formula for d(dy/dx)/dy is shown in attachment.
    Attached Thumbnails Attached Thumbnails d(dy/dt)/dy-derivatives-inverse-functions.jpg  
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