# Riddles! (argh they're pretty hard)

#### NewSc2

##### Diamond Member
Found it off a site that was slashdotted---

Riddles

I've gotten through most of the easys, but the medium ones and beyond get kind of difficult.

and.. when you need some comic relief (also elsewhere on the site)

How to be a Stereotypical Asian

#### thEnEuRoMancER

##### Golden Member
Damn, this one got me thinking...

We know that the derivative of x^2 with respect to x is 2x. However, what if we rewrite x^2 as the sum of x x's, and then take the derivative:

d/dx[ x2 ] = d/dx[ x + x + x + ... (x times) ]
= d/dx[x] + d/dx[x] + d/dx[x] ... (x times)
= 1 + 1 + 1 + ... (x times)
= x

This argument shows that the derivative of x2 with respect to x is actually x. So what's going on here?

Note: Most people with some math experience can show that some part of the argument is erroneous. As in simply, something doesn't follow. However, a full solution will explain why this argument attacks something that lies at the very heart of calculus itself, and that is what really explains why it's erroneous.

#### Whitecloak

##### Diamond Member
from first principles, we can prove that its wrong.

d/dx(function) = Limit y-> 0 [{function(x+y)-function(x)}/y]

writing it this we get

d/dx(x^2) = limit y->0 [{(x+y)+(x+y).....(x+y)times} - {x+x+x...x times}]/y

x x's cancel out leaving y x's & (x+y) y's

ie

= Limit y->0 [y.x+(x+y).y]/y = Limit y-> 0 [x + x + y]

= 2x

hence proved

The reasoning was wrong in the original argument. The assumption they made was that differential of x^2 = sum of differentials of x x's