Is the product rule broken?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Is the product rule broken?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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
  • phi
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.
Ahh ok I guess that makes sense.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 \(\Delta y\to0\) \[\frac{dA}{dx}=y+x\frac{dy}{dx}+0\] \[dA=ydx+xdy\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question