> Tell me one thing ... if f(x) = 5x,

> what is f'(x) when x = 100 ? 5 right ??

Right.

> and what about x = 200 ? again 5 right ??

Right.

> So, can you say 100 = 200 ?? NO.

I never said it was.

The way I defined f(x) was as follows: x+x+x...(x times). This is shorthand, and probably makes more sense in the longhand version:

(1) f(x)=f1(x)+f2(x)+f3(x)+...+fN(x), where fp(x)=x (p=1..N)

(2) x=N.

So where N=3, f(x)=x+x+x; where N=5 f(x)=x+x+x+x+x; where N=7 f(x)=x+x+x+x+x+x+x.

Therefore where N=3, f'(x)=1+1+1=3 which is independent of x. So it's still 3 when x=3. And for all N, f'(x)=N when x=N. So for all N, f'(x)=x. Which is what I said in my first post.

So if you still disagree with this, please provide JUST ONE value of N for which this DOES NOT hold. Should be easy to find.

> what is f'(x) when x = 100 ? 5 right ??

Right.

> and what about x = 200 ? again 5 right ??

Right.

> So, can you say 100 = 200 ?? NO.

I never said it was.

The way I defined f(x) was as follows: x+x+x...(x times). This is shorthand, and probably makes more sense in the longhand version:

(1) f(x)=f1(x)+f2(x)+f3(x)+...+fN(x), where fp(x)=x (p=1..N)

(2) x=N.

So where N=3, f(x)=x+x+x; where N=5 f(x)=x+x+x+x+x; where N=7 f(x)=x+x+x+x+x+x+x.

Therefore where N=3, f'(x)=1+1+1=3 which is independent of x. So it's still 3 when x=3. And for all N, f'(x)=N when x=N. So for all N, f'(x)=x. Which is what I said in my first post.

So if you still disagree with this, please provide JUST ONE value of N for which this DOES NOT hold. Should be easy to find.