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    1/(dy/dx) = dx/dy
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    They are reciprocals.
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    dy/dx is 1/dx/dy. is that what you meant?
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    As the others have said, provided that \dfrac{dy}{dx} \ne 0. Whenever \dfrac{dy}{dx} = 0, \dfrac{dx}{dy} isn't defined. And likewise, whenever \dfrac{dx}{dy}=0, \dfrac{dy}{dx} isn't defined.
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    one is the reciprocal of the other
    ie dx/dy = 1/ (dy/dx)

    for example
    if y=ax^2 + bx +c
    then differentiating with respect to x gives
    dy/dx = 2ax + b
    but differentiating with respect to y gives
    1=2ax(dx/dy)+b(dx/dy)
    dx/dy = 1/(2ax+b)
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    (Original post by Mr M)
    They are reciprocals.
    Indeed. :yep:

    They're just each other flipped over, if that makes more sense, OP. :dontknow:
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    (Original post by abbii)
    one is the inverse of the other
    Are you sure about this?
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    (Original post by Mr M)
    Are you sure about this?
    i meant it in the sence that they are recipicols
    i was wrong thinking about it
    the inverse to dy/dx would be the integral of y with respect to x
    i will edit earlier post
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    so is it dy/dx = 1/(dx/dy)
    or
    1/(dy/dx)=dx/dy
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    (Original post by uxa595)
    so is it dy/dx = 1/(dx/dy)
    or
    1/(dy/dx)=dx/dy
    both
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    (Original post by uxa595)
    so is it dy/dx = 1/(dx/dy)
    or
    1/(dy/dx)=dx/dy
    What's the difference?
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    (Original post by abbii)
    i meant it in the sence that they are recipicols
    i was wrong thinking about it
    the inverse to dy/dx would be the integral of y with respect to x
    i will edit earlier post
    Strictly speaking that's not true, although it is for A-level purposes, so to avoid confusion I won't say any more. (But it's good to know that when you say that you're doing a bad, bad, evil, terrible thing).

    (Original post by uxa595)
    so is it dy/dx = 1/(dx/dy)
    or
    1/(dy/dx)=dx/dy
    Both, because a = \dfrac{1}{b} \iff b = \dfrac{1}{a}.
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    (Original post by nuodai)
    Strictly speaking that's not true, although it is for A-level purposes, so to avoid confusion I won't say any more. (But it's good to know that when you say that you're doing a bad, bad, evil, terrible thing).
    I don't think you'd be confusing anyone at A-level particularly much by giving a brief outline of the problem. Pathological cases aside, the problem is simply that, if you take a function f, integrate it, then differentiate the result, you get f back; however, if you take a function g, differentiate it, then integrate the result, you get g + (some constant) - so, in fact, any function that has an integral has other integrals too by just tweaking the constant. They're essentially the same, but not quite. So integration and differentiation are almost inverses in those cases, but not quite.

    If anyone wonders what I mean by "pathological cases", well, for example... some functions just can't be differentiated everywhere (e.g. |x|), and likewise some functions can't be integrated everywhere (though coming up with an example of those requires a bit more creativity - imagine a function that fluctuates between positive and negative so erratically that its integral can't possibly converge), so it doesn't make sense to start comparing the derivative of the integral and "the" integral of the derivative.
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    I think its this:

    dy/dx = rate of change in y in relation to x

    dx/dy = rate of change of x in relation to y

    someone check please
 
 
 
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