Anonymous at Thu, 28 Mar 2024 22:13:50 UTC No. 16102103

butt x is a constant in your gayquation, so zero derivative is correct

Anonymous at Thu, 28 Mar 2024 22:16:00 UTC No. 16102104

oh, i didn't see the "x times". so really your equation is x = x so dx/dx = 1

Anonymous at Thu, 28 Mar 2024 22:30:09 UTC No. 16102128

I'm a graduate student and I can't prove this wrong

Anonymous at Thu, 28 Mar 2024 22:38:59 UTC No. 16102143

What if x = sqrt(2)
how does 1 + 1 + ... + 1 = sqrt(2)?
You need to properly define this process.

Anonymous at Thu, 28 Mar 2024 22:43:56 UTC No. 16102153

>>16102099
>>16102143
OP IN SHAMBLES

btw, dx/dx would just be 0/0, since x is unchanging.

Anonymous at Thu, 28 Mar 2024 22:44:46 UTC No. 16102154

>>16102153
0/0 = 1 btw

Anonymous at Thu, 28 Mar 2024 22:46:04 UTC No. 16102158

What if x = 0.999...?

Anonymous at Thu, 28 Mar 2024 22:52:07 UTC No. 16102172

>>16102099
That's a discrete function. Use the difference quotient

Anonymous at Thu, 28 Mar 2024 23:13:15 UTC No. 16102215

>>16102099
[math]
\frac{d}{dx} kx = \frac{d}{dx} x + \frac{d}{dx} x + . . . + \frac{d}{dx} x
[/math]
This is what occurs k times.

Anonymous at Fri, 29 Mar 2024 01:05:39 UTC No. 16102439

>>16102143
sqrr(2) clearly doesn't exist

Anonymous at Fri, 29 Mar 2024 01:09:46 UTC No. 16102447

>>16102439
Based, I fucking kneel