> When you write f(x) = x+x+x .. (x times) .. +x , what do you mean by 'x times' ??

Seems a fair question. If x=3 then f(x)=x+x+x. You see there are 3 x's, because x=3.

Similarly if x=7 then f(x)=x+x+x+x+x+x+x, and there are 7 x's because x=7.

So we have f(x)=f1(x)+f2(x)+f3(x)+...+fN(x), where f1(x)=f2(x)=..=fN(x)=x and N=x, if that makes it any clearer.

> Now, when you differentiate f(x) : [[ D(h(x)) = Derivative of h(x) w.r.t. x ]]

> you do it like f'(x) = D(x) + D(x) + ..(x times).. + D(x) which is invalid.

Why is it invalid? According to

http://mathworld.wolfram.com/Derivative.html "Derivatives of sums are equal to the sum of derivatives" (exact quote), see equation 36: [f(x)+...+h(x)]' = f'(x)+...+h'(x).

> Note that, I have used the multiplication formula above

Yes, I saw that, and that's what I think mayjune may have done too. But the product rule (equation 38 on the same page) relates to a(x).b(x), not a(x)+b(x), and so doesn't apply, because my f(x) is a *sum* of functions of x, not a *product* of functions of x.

Currently both proposed answers effectively declare the question invalid, but the question is not invalid and so there is no correct answer so far. Anyone else fancy a go? I won't add further clarifications, as I think the question as stated was quite clear. I get notifications of answers to this thread, so when I see the correct answer I will confirm this fairly quickly (during UK daylight hours).