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FreeTrader

  • 3 years ago

ALGEBRA in Calculus proof... Can you help me understand the basis for bringing delta x upstairs into only the numerator in this step of a problem?

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  1. FreeTrader
    • 3 years ago
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    \[1/\Delta x * (\Delta u)v - u(\Delta v)/(v + \Delta v)v\] \[((\Delta u / \Delta x)v - u(\Delta v/\Delta x))/(v + \Delta v)v\]

  2. FreeTrader
    • 3 years ago
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    This is from Lecture Session 10 of MIT OCW Math 18.01 Scholar - Single Variable Calculus.

  3. FreeTrader
    • 3 years ago
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    The lecture and lecture notes show what I typed above. I don't understand why the Delta x would not have to be multiplied against the (v + Delta v)v term in the denominator.

  4. FreeTrader
    • 3 years ago
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    How does the Delta x term get into the numerator of the numerator?

  5. TuringTest
    • 3 years ago
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    \[{{1\overΔx}(Δu)v−u(Δv)\over(v+Δv)v} \to {{Δu \overΔx}v−u{Δv \overΔx}\over(v+Δv)v}\]is the step?

  6. FreeTrader
    • 3 years ago
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    Almost. The 1/Delta x is a separate fraction. It started out as simply Delta x under everything else.

  7. FreeTrader
    • 3 years ago
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    And then ended up as you have on the right hand side.

  8. FreeTrader
    • 3 years ago
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    see attached for more clarity. It's near the bottom third of the page...

  9. TuringTest
    • 3 years ago
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    oh this is simple: multiply out the bottom by delta x then divide the top and bottom by delta x much easier when I can see it!

  10. TuringTest
    • 3 years ago
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    or divide top and bottom of \[{1\over \Delta x}\] by \[\Delta x\]you get\[{{1\over \Delta x}\over1}{(\Delta u)v-u (\Delta v)\over(v+\Delta v)v}\]

  11. FreeTrader
    • 3 years ago
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    I need to go to Algebra Tricks Boot Camp. Thanks.

  12. FreeTrader
    • 3 years ago
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    I still think that is really funky. I will need to do some test problems to convince my rock-like brain this works just like you show it does.

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