Where is the error in this derivative example?

Question

x^2=x+x+x+...(x times)...+x
Let's take a derivative d/dx
2x=1+1+1+...(x times)+...1
so that
2x=x
Why derivative on the lefthand side is not equal to the derivative on righthand side

Answer 1

I had the same misinterpretation as como at first; but notice that x*x, by our traditional explanation of multiplication, is x+x+...+x, where there are x terms in the sum. As BNP pointed out, this doesn't work too well when we generalize to real numbers.

However, I think the main problem here is that your sum depends on x. Taking the derivative of each side, you say the RHS is
1+1+...+1, with x terms. But we just took the derivative with respect to x! So saying there are "x terms" doesn't work out anymore. Consider the following analogy, using the same traditional explanation of multiplication:

x = 1+1+...+1 (x terms).
Take derivative wrt x:
1 = 0+0+...+0 (x terms).

Again, the problem shows up in that we have "x terms", which we didn't bother to change when we differentiated wrt x.

Answer 2

I think the problem comes because the right hand side is not sensible for non-integers. For instance, if x is 1.5, how do you have a discretized sum of x+x+...?
Differentiation requires real numbers, not integers.

Edit: Ah, just had a moment of insight. To write it as a sum, you would have to write it as Integral(from 0 to x) of x dy. (yes, I really mean dy).

Answer 3

the error is in first line: x^2<>x+x+x+...(x times)...+x, but x^2= x *x

Answer 4

x ² = (x) (x)
Taking derivative of each side:-
2x = (1)(x) + (1)(x)
2x = x + x
2x = 2x (as required)
Your first line is wrong in that x² = (x) (x) and not x + x + x + x + x------as you suggest.