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I proved differential of x^2 is x!

Anonymous Sun 12 Jul 17:26:41 2020 No.11893650 View Reply Original Report

Quoted By: >>11893675 >>11893680 >>11893747 >>11893796 >>11893803 >>11893892 >>11893911

Q.E.D

Anonymous

Anonymous Sun 12 Jul 2020 17:34:10 No.11893675 Report

Quoted By: >>11893795

>>11893650
1/7 = 1 + ... + 1 1/7-times. right

Anonymous

Anonymous Sun 12 Jul 2020 17:35:48 No.11893680 Report

Quoted By: >>11893795

>>11893650
retarded you are not in $\mathbb{N}$

Anonymous

Anonymous Sun 12 Jul 2020 17:54:23 No.11893735 Report

Quoted By:

$ f(x)=x^2 \\ \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \dfrac{(x+h)^2 -x^2}{h} \\ \displaystyle = \lim_{h \to 0} \dfrac{x^2+2 xh+h^2 -x^2}{h} = \lim_{h \to 0} ~ 2x+h = 2x $

Anonymous

Anonymous Sun 12 Jul 2020 17:55:26 No.11893739 Report

Quoted By: >>11893843

$ f(x)=x^2 \\ \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \dfrac{(x+h)^2 -x^2}{h} \\ \displaystyle = \lim_{h \to 0} \dfrac{x^2+2 xh+h^2 -x^2}{h} = \lim_{h \to 0} ~ 2x+h = 2x $

Anonymous

Anonymous Sun 12 Jul 2020 17:57:08 No.11893747 Report

Quoted By:

>>11893650
On second thought, perhaps Bourbaki was right....

Anonymous

Anonymous Sun 12 Jul 2020 18:11:06 No.11893795 Report

Quoted By:

>>11893675
>>11893680
retards. just add a term to make up the difference between integers and reals, so then the derivative will contain some constant term but will be x for whole numbers.

Anonymous

Anonymous Sun 12 Jul 2020 18:11:11 No.11893796 Report

Quoted By:

>>11893650
So the derivative of f(x)=x is 0? Nice

Anonymous

Anonymous Sun 12 Jul 2020 18:14:44 No.11893803 Report

Quoted By:

>>11893650
That's not x! though

Anonymous

Anonymous Sun 12 Jul 2020 18:31:02 No.11893843 Report

Quoted By: >>11893880 >>11893907

>>11893739
How do you generalise that for $x^n$ ?

Anonymous

Anonymous Sun 12 Jul 2020 18:40:32 No.11893880 Report

Quoted By: >>11893897

>>11893843
use binomial theorme

Anonymous

Anonymous Sun 12 Jul 2020 18:43:17 No.11893892 Report

Quoted By:

>>11893650
You didn't differentiate the x at the top of the sideways curly brace. You need to use a Leibniz rule to do so, in which case you get x 1's summed, and 1 x summed, which is 2x.

Anonymous

Anonymous Sun 12 Jul 2020 18:44:15 No.11893897 Report

Quoted By:

>>11893880
But how would that help if n is not an integer exponent?

Anonymous

Anonymous Sun 12 Jul 2020 18:47:15 No.11893907 Report

Quoted By:

>>11893843
product rule

Anonymous

Anonymous Sun 12 Jul 2020 18:49:09 No.11893911 Report

Quoted By:

>>11893650

If you wrote this as a sum, the dependent variable would appear in the limits. When you differentiate you neglect it.

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