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)