D(x^2) = x

We usually denote derivative like this: f'(x) where f is a function and x the variable, right.
Let me from here on use the following notation for simplicity: D(f). If f(x) = x^2 then I write directly D(x^2) for its derivative, may I do this?

Okay, everyone knows that D(x^2) = 2*x, right. There is no argue about this.

So if I 'prove' that D(x^2) = x it should obviously be some error in the 'proof'.

But look at this:
D(x^2) =
= D(x + x + x + ...) {with x number of terms} =
= D(x) + D(x) + D(x) + ... {with x number of terms} =
= 1 + 1 + 1 + ... {with x number of terms} =
= x
Hence D(x^2) = x ???

This 'proof' is obviously wrong somewhere, but where is the error to be found?

Interesting. There seems to something funny with the assumption that X^2 = X + X + X .... (for X terms). I see that this works, sort of, but what if x was a measurement that had units attached, like meters. Then (5m)^2 is not equal to 5m + 5m + 5m + ... The units are inconsistent. And how would you have 5m terms?

Originally posted by FabianFnas
We usually denote derivative like this: f'(x) where f is a function and x the variable, right.
Let me from here on use the following notation for simplicity: D(f). If f(x) = x^2 then I write directly D(x^2) for its derivative, may I do this?

Okay, everyone knows that D(x^2) = 2*x, right. There is no argue about this.

So if I 'prove' that D(x^2) = x ...[text shortened]... = x ???

This 'proof' is obviously wrong somewhere, but where is the error to be found?

That's because you don't derive the x number of terms.

Originally posted by Palynka
That's because you don't derive the x number of terms.

No? I did, didn't I?

D(3*x) = D(x+x+x) = D(x) + D(x) + D(x) = 1 + 1 + 1 = 3 Right?
This could be written
D(three times x) = D(three terms of x) = D(x) three times = 1 three times = 3 Right?

So of course we can derive x number of times.

Originally posted by FabianFnas
No? I did, didn't I?

D(3*x) = D(x+x+x) = D(x) + D(x) + D(x) = 1 + 1 + 1 = 3 Right?
This could be written
D(three times x) = D(three terms of x) = D(x) three times = 1 three times = 3 Right?

So of course we can derive x number of times.

Sorry if I wasn't clear, I meant that the x number of times is also variable (unlike 3 in your example).

Originally posted by Palynka
Sorry if I wasn't clear, I meant that the x number of times is also variable (unlike 3 in your example).

I'm not sure myself where the error is to be found in my 'proof', though I understand that there must be an error somwhere.

So, this is the error, that x is a variable in itself, and not a constant? Then I thank you, this has boggled my mind for a while...

Originally posted by FabianFnas
I'm not sure myself where the error is to be found in my 'proof', though I understand that there must be an error somwhere.

So, this is the error, that x is a variable in itself, and not a constant? Then I thank you, this has boggled my mind for a while...

Yes, D(x.x)=D(x+x.+.. (x times)) is different than x.D(x)=D(x)+D(x)+...(x times).

If you look at it as a derivative of a product, you'll have D(x).x+x.D(x)=2x

Edit - Intuitively, you can see that a change in x, has a double effect on x+x..(x times): first, each element of the sum changes and secondly the number of elements also changes.

Palynka is correct, so I won't repeat him.

x^2=x+x+x+...+x (x times) only if x is a non-negative integer, but to define the derivative x must be continuous.

Tahnk you Palynka for your explanation.
When I present the problem ot som fellow math student, they just shake their heads.

You assume that the D() has a distributive property with addition when such is not the case, nor does it make any sense.

D() takes a function as it's argument, and as such, operations which require specific values cannot be used, rendering distribution meaningless.

Now what is true is this:

Suppose h(x) = f(x) + g(x), h'(x) = f'(x) + g'(x).