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# Funny thing about x/0 Watch

1. (I do A2, if I've made a glaring mistake here like I believe I have please point it out)

Now as I'm sure everyone is aware, any number other than zero divided by zero has no value, however something odd happens when I differentiate the function

y=x/0

U=x V=0

y=U/V

Quotient rule

dy/dx=(V(du/dx)-U(dv/dx))/V^2

Substitute x and zero back in

dy/dx=(0-0)/0

Here's the weird part, 0/0 can be anything, so for any value of x, even though y is undefined, dy/dx is equal to every single number at the same time... How very... odd...

I await patronising explanations of some fundamental misunderstanding

Edit: The obvious absurdity here

dy/dx=0/0

y=Integral 0/0 dx

(Multiplying anything by zero is zero so I can shove in an x I think)

y= integral 0/0x dx

y=ln0x=ln0 (I don't think I can do that)

x/0(Is identical to)ln0

0/0=ln0

ln0 is defined as all numbers

x/0 satisfies all values
2. Dividing by zero glitches maths. If you allow dividing by zero, pi = 42 and 3=eleventy twelve.

Hence the odd results.

(Disclaimer: I'm only an A2 student, hence the simplistic reply. I await some postgrad to use axiomatic set theory to prove why this happens)

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3. Before I'm slam dunked can I suggest that it's probably because the proof of the quotient rule assumes you are not trying to divide by zero?
4. If maths is inconsistent then absolutely everything is true. Since x/0 being well defined would make maths inconsistent, you can use it to prove absolutely anything.
5. A function has to be continuous to be differentiable at a point ergo you cannot differentiate x/0 as its not a continuous function.
6. (Original post by MM04926412)
Before I'm slam dunked can I suggest that it's probably because the proof of the quotient rule assumes you are not trying to divide by zero?
Try using product rule. Ie x(0)^-1 ha.

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7. (Original post by Mathlover123)
A function has to be continuous to be differentiable at a point ergo you cannot differentiate x/0 as its not a continuous function.
It isn't even a function.
8. (Original post by physicsmaths)
Try using product rule. Ie x(0)^-1 ha.

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You realise that the quotient rule is just the product rule?
9. (Original post by TenOfThem)
You realise that the quotient rule is just the product rule?
obviously as thats why i rewrote his thing so he could use the product rule

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10. (Original post by TenOfThem)
You realise that the quotient rule is just the product rule?
y=uv^-1 and differentiating gives quotient rule.
i wrote ha at then end for a reason lol.

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11. (Original post by physicsmaths)
obviously as thats why i rewrote his thing so he could use the product rule

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(Original post by physicsmaths)
y=uv^-1 and differentiating gives quotient rule.
i wrote ha at then end for a reason lol.

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I guess you were trying to make a point that I have failed to grasp
12. (Original post by TenOfThem)
I guess you were trying to make a point that I have failed to grasp
Lol no i was joking and wanted tos ee if he would realise what u said. thats why i wrote ha at the end lol.

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13. (Original post by james22)
It isn't even a function.
Yeah I guess that's just as fair a point to differentiate a function you must at least have a function.
14. (Original post by physicsmaths)
Lol no i was joking and wanted tos ee if he would realise what u said. thats why i wrote ha at the end lol.
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Updated: January 9, 2015
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