calculus I question?

The stepsin your calculation:
Step [1]: (1 + 1 + 1 + …… + 1) x = x²
Step [2]: Take derivative on both sides:  1 + 1 + 1 + …… + 1 = 2x
Step [3]: x = 2x ? …… [3]

The reason of wrong calculation:
From step [1] to step [2]: (1 + 1 + 1 + …… + 1) was treated as a constant.
From step [2] to step [3]: (1 + 1 + 1 + …… + 1) was treated as variable x.

Actually, (1 + 1 + 1 + …… + 1) = x and thus it a variable.
The derivative on the left side is wrong.

The only thing wrong is that you do not even know the basic axioms of arithmetic, so how you reckon to do calculus is a mystery 

(x+x....+x), x times= x^2 ?  NO IT DOESN'T.

Learn the basics and start over.

deriv. of x^3  =  3x^2
derv. of x^5  =  5x^4

if we take derivative of both sides we get (1+1+1..+1) x times = 2x
 ...  as you said ...  this is a derivative, so what is your math. reasoning for setting this equal to the original equation?  ...  there is NONE.
  .but (1+1+1..).x times =x   <<<  this is the orig. 'equation' 
 ....  why are you setting the original equal to a derivative???  ==  makes no math sense.

  ALSO
   (2 + 2)  =  2^2  ....  and  (3 + 3 + 3)  =  3^2
  .but (1+1+1..).x times =x,  in this case x times means to sum 1 exactly 1 time so it is really (1 + nothing)  = 1^2 = 1
    so     1  =  1^2   ....  correct
     deriv.  ???  left .. deriv of  1  =  0  <<<  deriv of a constant is 0
     deiv. of right  =  ... deriv of 1^2??  still a constant .. =  0  not 2x .. this equation has no x variable

what is wrong ?
Is that you really don't have two equation which are 
equal? 
You have one equation which is equal to another 
only when x has a specific value.  

you are  
if 
f(x) =ax     
g(x) = x^2      

f(a) = a^2 
g(a) = a^2   
they are equal  only when   the x value  = a  
That is only that have shown

ax=x^2  
x^2 -ax  = 0 
x(x-a) = 0 
so this is only true when x =0 or x = a  

they are not equal all the time.     

since they are only equal when x =a  or x = 0 
then why should their derivatives be equal.