Nope, that's where you go wrong.
f'(x) = x as long as N is a number independent of x.

When N is independent of x, f'(x) = N.
See, f(x) = x + x + ..(N times).. + x = Nx
So, f'(x) = N.

I think an example might make things a bit clearer.
Let us define f(x) = x + x + x = 3x.
So f'(x) = 3.
Now, f'(x) = 3 is completely independent of x.
So, f'(x) is not gonna change whether x=3,4,5 or anything.
Universally, if f(x) = 3x, f'(x) is always equal to 3, whatever x be.

D(Nx) = N. But, D(x.x) != x.

Tell me one thing ... if f(x) = 5x,
what is f'(x) when x = 100 ? 5 right ??
and what about x = 200 ? again 5 right ??
So, can you say 100 = 200 ?? NO.

When you write f(x) = Nx, your func is a linear one with constant slope -- so has a constant derivative at all values of x.
When you write f(x) = x times x, it depends on x twice. So, it is basically quadratic in nature. It's slope varies with x.

You are accepting things in the wrong way .. I mean try to feel the Maths behind, not the formula.
Formulas may deceive you, but logic never does.

And, if my explanation still doesn't satisfy you.. why not ask someone who you trust ??