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Thread: 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 $\displaystyle e^t$ for $\displaystyle x$. It followed that $\displaystyle x=e^t\Leftrightarrow t=ln|x|.$ Then $\displaystyle dt=\frac{1}{x}\,dx$ or simply $\displaystyle \frac{dt}{dx}=\frac{1}{x}$. But what about $\displaystyle \frac{dx}{dt}$? Is it $\displaystyle x$ or $\displaystyle e^t$. I'm very confused about this because whenever we have a 1st order ODE let's say $\displaystyle y'=f(x)$ we change it to $\displaystyle \frac{dy}{dx}=f(x)$ and then multiply the RHS by $\displaystyle dx$ follows that $\displaystyle dy=f(x)\,dx$ 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 $\displaystyle e^t$ for $\displaystyle x$. It followed that $\displaystyle x=e^t\Leftrightarrow t=ln|x|.$ Then $\displaystyle dt=\frac{1}{x}\,dx$ or simply $\displaystyle \frac{dt}{dx}=\frac{1}{x}$. But what about $\displaystyle \frac{dx}{dt}$? Is it $\displaystyle x$ or $\displaystyle e^t$. I'm very confused about this because whenever we have a 1st order ODE let's say $\displaystyle y'=f(x)$ we change it to $\displaystyle \frac{dy}{dx}=f(x)$ and then multiply the RHS by $\displaystyle dx$ follows that $\displaystyle dy=f(x)\,dx$ as if the latter were a denominator!

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

    $\displaystyle since\ x=e^t$

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

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

    whereby integration is performed as anti-differentiation.

    $\displaystyle y\ '(x)=f(x)$

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

    $\displaystyle dy=f(x)dx$

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

    $\displaystyle since\ x=e^t$

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

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

    Thanks, well I am aware that $\displaystyle \frac{dy}{dx}$ is essentially a slope, but when we can treat $\displaystyle dx$ as a denominator, can we treat $\displaystyle dy$ 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 $\displaystyle x=e^t\frac{dx}{dt}=e^t$ but what about $\displaystyle \frac{dt}{dx}$?? Why isn't it just the inverse as in $\displaystyle frac{1}{x}$. 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 $\displaystyle \frac{dy}{dx}$ 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 $\displaystyle \frac{dx}{dt}=e^t$ and $\displaystyle \frac{dt}{dx}=\frac{1}{x}$ and not $\displaystyle e^{-t}$ ??
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  6. #6
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    It is exp(-t).

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

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