Differentiation problem

f(x) = x+x+x... (x times)
g(x) = x^2

Obviously f(x) = g(x), for example
if x=4 then f(x)=x+x+x+x = 4+4+4+4 = 16
if x=5 then f(x)=x+x+x+x+x = 5+5+5+5+5 = 25

Now let's differentiate both.

f'(x)=d/dx x + d/dx x + d/dx x... (x times)
and since d/dx x = 1, f'(x)=1+1+1... (x times) = x.

g'(x)=d/dx x^2 = 2x

Explain why f'(x) != g'(x).

You can contribute to $1 Competition

It is equal,
f(x)= x + x + x + x (x times) = x * x
Now,
f'(x) = (d/dx x) * x + x * (d/dx x)
f'(x) = 1 * x + x * 1
f'(x) = x + x = 2x = g'(x)

Is that mean that f'(X) and g'(X) are same in above case. As the question was to explain why f'(x) != g'(x).

Let me clarify that != is used in the C sense, i.e. DOES NOT EQUAL.
So the puzzles is to explain why f'(x) does not equal g'(x).

f'(x) != g'(x) because there is an error in the calculation of f'(x).

When you write f(x) = x+x+x .. (x times) .. +x ,
what do you mean by 'x times' ?? x is not something like 1, 2, 3 .. it's just a variable.
So, you are writing f(x) as sum of a variable, variable times.

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.
The formula D(nx) = nD(x) is valid ONLY for constant n, not a variable one.
So, f'(x) != x * D(x)

PS : The correct way will be to do it like :
f(x) = x + x + ..(x times).. + x
So, f'(x) = D(x) + D(x) + ..(x times).. + D(x) + x
Note that, I have used the multiplication formula above.
Multiplication formula : D( u(x) * v(x) ) = v(x)*D(u(x)) + u(x)*D(v(x))

So, f(x) = x + x + ..(x times).. + x
=> f'(x) = xD(x) + x
=> f'(x) = x+x = 2x = g'(x)

> 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).

Its already 24 hours and so the winner is the OP as per the rule

Let's have the answer then, this question was asked to me by my friend two years ago, he gave the same explanation me and saswat gave, lets see how you answer it xp

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.