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Math Help - Derivative confusion

  1. #1
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    Derivative confusion

    Hello everyone!

    I've lately came accross this "confusion" as I was solving a cauchy-euler equation that required a subtitution of e^t for x. It followed that x=e^t\Leftrightarrow t=ln|x|. Then dt=\frac{1}{x}\,dx or simply \frac{dt}{dx}=\frac{1}{x}. But what about \frac{dx}{dt}? Is it x or e^t. I'm very confused about this because whenever we have a 1st order ODE let's say y'=f(x) we change it to \frac{dy}{dx}=f(x) and then multiply the RHS by dx follows that dy=f(x)\,dx as if the latter were a denominator!

    If someone clears that out for me I would be very greatful!
    Thanks!!
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  2. #2
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    Quote Originally Posted by rebghb View Post
    Hello everyone!

    I've lately came accross this "confusion" as I was solving a cauchy-euler equation that required a subtitution of e^t for x. It followed that x=e^t\Leftrightarrow t=ln|x|. Then dt=\frac{1}{x}\,dx or simply \frac{dt}{dx}=\frac{1}{x}. But what about \frac{dx}{dt}? Is it x or e^t. I'm very confused about this because whenever we have a 1st order ODE let's say y'=f(x) we change it to \frac{dy}{dx}=f(x) and then multiply the RHS by dx follows that dy=f(x)\,dx as if the latter were a denominator!

    If someone clears that out for me I would be very greatful!
    Thanks!!
    Hi rebghb,

    since\ x=e^t

    \frac{dx}{dt}=e^t=x=\frac{1}{\left[\frac{dt}{dx}\right]}

    As \frac{dy}{dx} is a ratio at any given x, being the slope of the tangent to the graph, you may treat dx as a denominator,

    whereby integration is performed as anti-differentiation.

    y\ '(x)=f(x)

    \frac{dy}{dx}=f(x)

    dy=f(x)dx

    \int{dy}=y=\int{f(x)}dx
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  3. #3
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    [quote=Archie Meade;498895]Hi rebghb,

    since\ x=e^t

    \frac{dx}{dt}=e^t=x=\frac{1}{\left[\frac{dt}{dx}\right]}

    As \frac{dy}{dx} is a ratio at any given x, being the slope of the tangent to the graph, you may treat dx as a denominator,
    quote]

    Thanks, well I am aware that \frac{dy}{dx} is essentially a slope, but when we can treat dx as a denominator, can we treat dy as a numerator? (if the bottom part is the denominator then what about the upper part). I got your explanation using the definition of an integral (well not exactly definition) but if x=e^t\frac{dx}{dt}=e^t but what about \frac{dt}{dx}?? Why isn't it just the inverse as in frac{1}{x}. I'm lost because noone has made this clear, deriving has always been a routine
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  4. #4
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    Yes. While \frac{dy}{dx} is NOT a fraction, it is a limit of one and so we can typically treat it like a fraction. That is one of the reasons for defining the "differential", dx and dy with dy= f'(x)dx. That allows us to formally treat dy/dx as a fraction.
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  5. #5
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    Okay I'll ask this bluntly, this will sound stupid: Why is it thay \frac{dx}{dt}=e^t and \frac{dt}{dx}=\frac{1}{x} and not e^{-t} ??
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  6. #6
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    It is exp(-t).

    \frac{dx}{dt}=e^t=x

    \frac{dt}{dx}=\frac{1}{x}=\frac{1}{e^t}=e^{-t}=\frac{1}{\frac{dx}{dt}}
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  7. #7
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    aaaah! Thanks!
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