Mathematics is simple but tricky...........look at this.........u will understand......how tricky maths was?

As we know that the derivative of x² , with respect to x , is 2x

i.e., d/dx(x²)=2x

However, suppose we write x²  as the sum of x ‘s written up x times..

i.e x²=x+x+x+…+x times

1²=1

2²=2+2

3²=3+3+3……………………

……………………………….

x²=x+x+x+x+x+…………….+x times

let us take f(x)=x²=x+x+x+x+……….x times

now. F¹(x)= d/dx (x²)=d/dx(x+x+x+……..+x times)

i.e F¹(x)= d/dx(x)+ d/dx(x)+ d/dx(x)+ ………x times

             =1+1+1……….+x times

            = is actually x, not 2x..

Where is the error?

Error: x²

will equal to x+x+x+…+x x times only when x is a positive integer (i.e., x∈Z+. But for the differentiation, we define a function as the function of a real variable. Therefore, as x is a real number, there arises a domain R−Z+ where the statement x2=x+x+x+…+x x times fails.

And since, the expansion  x²≠x+x+x+…+x x times  for x∈R

, the respective differentiations will not be equal to each other.

Then how can x²expanded in such a way?

If x is a positive integer:

x²=x+x+x+…+x x times

But when when x is an arbitrary real number >0, then x can be written as the sum of it’s greatest integer function [x] and fractional part function {x}. 

Therefore, x²=[x]⋅x+x⋅x

x²=(x+x+…+x) [x]times +x⋅x

So, we can now correct the fallacy by changing the solution steps to:

x²=x[x]+x{x}

d/dx[x²]=d/dx(x[x]+x{x})

(differentiation by part)

=1⋅[x]+x⋅[x]′+1⋅{x}+x⋅{x}′

since d/dx(x)=x′=1

and [x]’ & {x}’ represent differentiation of each with respect to x.

=[x]+{x}+x([x]′+{x}′)

=x+x(x′)=x+x=2x