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xpi0t0s 9Aug2009 15:32

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).

shabbir 9Aug2009 17:03

Re: Differentiation problem
You can contribute to $1 Competition

mayjune 9Aug2009 18:26

Re: Differentiation problem
It is equal,
f(x)= x + x + x + x (x times) = x * x
f'(x) = (d/dx x) * x + x * (d/dx x)
f'(x) = 1 * x + x * 1
f'(x) = x + x = 2x = g'(x)

Kshiteej 10Aug2009 11:12

Re: Differentiation problem
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).

xpi0t0s 10Aug2009 11:59

Re: Differentiation problem
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).

SaswatPadhi 10Aug2009 13:33

Re: Differentiation problem
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) :)

xpi0t0s 10Aug2009 15:00

Re: Differentiation problem
> 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).

shabbir 10Aug2009 16:05

Re: Differentiation problem
Its already 24 hours and so the winner is the OP as per the rule :D

mayjune 10Aug2009 16:08

Re: Differentiation problem
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 :)

xpi0t0s 10Aug2009 17:17

Re: Differentiation problem
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

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]

If f(x)=f1(x)+f2(x)+f3(x)+...+fN(x) then by equation 36:
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.

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