Given 'x' not equal to 0, let us consider the follwoing relation:

x + x + x + .... +x (added 'x' times) = x²

Differentiating both sides with respect to x, we get:

1 + 1 + 1 + 1 + .... + 1 ('x' times) = 2x

(Since the derivative of x² with respect to 'x' is 2x).

So we now have:

x = 2x

Cancelling 'x' from both sides, we have:

1 = 2

Now the very obvious question follows:

Where is the flaw ?

(In reply to

Additional Consideration by K Sengupta)

In general, if:

h(x) = (x^m+x^m+ .....+x^m) (x^n times)

then, in terms of product rule, we obtain:

h'(x)

= [m*x^(m-1)+m*x^(m-1) + .....+ m*x^(m-1)][x^n times] +

[x^m+x^m+ .....+x^m) [n*x^(n -1)times]

= x^(m-1)*X^(n-1)(mx+nx)

= (m+n)*x^(m+n-1)

This is indeed true, since:

d/dx[x^(m+n)]

= (m+n)*x^(m+n-1)