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Empty
 one year ago
Is the product rule broken?
Empty
 one year ago
Is the product rule broken?

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Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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

phi
 one year ago
Best ResponseYou've already chosen the best response.0dx dy is "second order" and goes to zero somewhere in all of this there is a limit, that causes that term to drop out.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahh ok I guess that makes sense.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1instead 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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