OK. To solve this you have to go back to first principles of differentiation.
See equation 6 at http://mathworld.wolfram.com/Derivative.html
f'(x)=lim(h->0) [f(x+h)-f(x)]/h

For f(x)=x, f'(x)=lim(h->0) [f(x+h)-f(x)]/h
=lim(h->0) [(x+h)-x]/h
=lim(h->0) [h]/h
=1

For g(x)=x^2, g'(x)=lim(h->0) [g(x+h)-g(x)]/h
=lim(h->0) [(x+h)^2-x^2]/h
=lim(h->0) [(x^2+2hx+h^2)-x^2]/h
=lim(h->0) [2hx+h^2]/h
=lim(h->0) [2x+h]
=2x

If f(x)=f1(x)+f2(x)+f3(x)+...+fN(x) then by equation 36:
f'(x)=f1'(x)+f2'(x)+f3'(x)+...+fN'(x)
So if fp(x)=x (where p=1..N) and fp'(x)=1 then
f'(x)=1+1+1+...1 (N times)
so f'(x)=N.
If N=x then f'(x)=x.

So for example let's say N=4.
Then f(x)=x+x+x+x, or f(x)=4x if you prefer.
Then f'(x)=1+1+1+1=4
This is constant for all x, even when x=4.

g(x)=x^2, so g'(x)=2x, which is 8 when x=4.

Since 4!=8, f'(x)!=g'(x), and this is true for all x and all N.