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theEric Group Title

Hi! I'm looking at an example from my book and I'm hoping somebody can explain this. It's in the first chapter, but I don't want to skip out on anything. Thanks! Here is the issue: My book says that \(\dfrac{dp/dt}{p-900}=\dfrac{1}{2}\) where \(p\neq 900\). Theennn... "By the chain rule the left side of [that equation] is the derivative of \(\ln\left|p-900\right|\) with respect to \(t\), so we have \(\dfrac{d}{dt} \ln\left| p-900\right|=\dfrac{1}{2}\)" My issue is that I'm missing something. I see that \(\cancel{\dfrac{d}{dt}\dfrac{1}{p-900}= \ln \left| p-900\right|}\).

  • 11 months ago
  • 11 months ago

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  1. theEric Group Title
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    I know the chain rule, like if \(f=f(g(t))\) then \(\dfrac{df}{dt}=\dfrac{df}{dg}\dfrac{dg}{dt}\). I think I'm just getting confused from looking at this too much...

    • 11 months ago
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    I crossed that out because it's really \(\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left| p-900\right|\)

    • 11 months ago
  3. theEric Group Title
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    And so comes my problem... The book says that \(\dfrac{dp/dt}{p-900}=\dfrac{1}{2}\) and \(\dfrac{d}{dt}\ln\left|p-900\right| = \dfrac{1}{2}\) Which means that \(\dfrac{dp/dt}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right| \) And I think that \(\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right| \) I can't be right unless \(\dfrac{dp/dt}{p-900}=\dfrac{1}{p-900}\). This would imply that \(dp/dt=1\), which isn't necessarily true for this problem, that I see.

    • 11 months ago
  4. theEric Group Title
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    I said that I think \(\dfrac{1}{p-900}=\dfrac{d}{dt}\ln\left|p-900\right|\), but really \(\dfrac{1}{p-900}=\dfrac{d}{dp}\ln\left|p-900\right|\).

    • 11 months ago
  5. theEric Group Title
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    So then.... \(\dfrac{dp/dt}{p-900}=\dfrac{dp}{dt}\dfrac{d}{dp}\ln\left|p-900\right|\)

    • 11 months ago
  6. theEric Group Title
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    Ohhh....

    • 11 months ago
  7. theEric Group Title
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    \(\dfrac{dp}{dt}\dfrac{d}{dp}\longrightarrow\dfrac{d}{dt}\)

    • 11 months ago
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