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  • one year ago

Is the product rule broken?

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  1. Empty
    • one year ago
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    Ok obviously not, but something weird happened when I was considering a change of area of a square by an infinitesimal amount: |dw:1438695359737:dw| So infinitesimally changing the area of the square is that extra area there \(dA\) but if you add it up by considering the little changes in x and y separately you get that Geometrically two slender rectangles and a tiny corner: \(dA=ydx+xdy+dxdy\) But if you use the product rule you get: \(dA=d(xy)=xdy+ydx\) So where did the dxdy term go? Actually isn't \(dxdy=dA\)? I feel like I've confused myself here haha

  2. phi
    • one year ago
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    dx dy is "second order" and goes to zero somewhere in all of this there is a limit, that causes that term to drop out.

  3. Empty
    • one year ago
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    Ahh ok I guess that makes sense.

  4. Zarkon
    • one year ago
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    instead of \[dA=ydx+xdy+dxdy\] you should probably write \[\Delta A=y\Delta x+x\Delta y+\Delta x\Delta y\] then \[\frac{\Delta A}{\Delta x}=\frac{y\Delta x}{\Delta x}+\frac{x\Delta y}{\Delta x}+\frac{\Delta x\Delta y}{\Delta x}\] \[\frac{\Delta A}{\Delta x}=y+x\frac{\Delta y}{\Delta x}+\Delta y\] letting \(\Delta x\to0\) and therefore \(\Delta y\to0\) \[\frac{dA}{dx}=y+x\frac{dy}{dx}+0\] \[dA=ydx+xdy\]

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