# Differentiation problem

Discussion in '\$1 Daily Competition' started by xpi0t0s, Aug 9, 2009.

1. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Forgot to say one thing more --- this time I want a reply from you (xpi0t0s) about who is right ....
... I know you would try to find out who actually is.

2. ### xpi0t0sMentor

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I'll take that as a Yes, i.e. you agree so far.

(1) Do you agree that for all x, f'(x)=5?
(2) Do you agree that for x=5, f'(x)=5?

I anticipate Yes for both, but if at either point you say no, please explain why.

3. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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What is f(x) ?
If f(x) = x + x + .. (x times).. + x, then answer would be --

(1) No
(2) No

Explanation :
(1) For all x, f'(x) would not be some constant like 5, or 10.
It would be 2x.
(2) For x = 5, f'(x) would be 2*5 = 10.

(Very basic differentiation question.)

4. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Or, if you consider f(x) = x + x + x + x + x i.e. 5x then ...

(1) Yes
(2) Yes

because f'(x) = 5 which is independent of x.

5. ### xpi0t0sMentor

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Yes, that was the question.

Next step: if f(x)=x+x+x+x+x and x=5, do you agree or disagree that f'(x)=x?

6. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Yeah.
f(x) = 5x => f'(x) = 5 = x.

So ???

7. ### xpi0t0sMentor

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So, generalising this, if f(x)=x+x+x+... (N times) and x=N, do you agree or disagree that f'(x)=N?

8. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Yeah .. as long as N is a number like 2, 3 -9, 11.2 or any number ... f'(x) = N.

9. ### xpi0t0sMentor

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OK, so since f'(x)=N and N=x, f'(x) must equal x. QED.

10. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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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 ??

11. ### xpi0t0sMentor

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> Tell me one thing ... if f(x) = 5x,
> what is f'(x) when x = 100 ? 5 right ??

Right.

> and what about x = 200 ? again 5 right ??

Right.

> So, can you say 100 = 200 ?? NO.

I never said it was.

The way I defined f(x) was as follows: x+x+x...(x times). This is shorthand, and probably makes more sense in the longhand version:
(1) f(x)=f1(x)+f2(x)+f3(x)+...+fN(x), where fp(x)=x (p=1..N)
(2) x=N.

So where N=3, f(x)=x+x+x; where N=5 f(x)=x+x+x+x+x; where N=7 f(x)=x+x+x+x+x+x+x.
Therefore where N=3, f'(x)=1+1+1=3 which is independent of x. So it's still 3 when x=3. And for all N, f'(x)=N when x=N. So for all N, f'(x)=x. Which is what I said in my first post.

So if you still disagree with this, please provide JUST ONE value of N for which this DOES NOT hold. Should be easy to find.

12. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Don't you really get what I say, or you are pretending ????
I said it would hold for N and not for x, if N is a number.

If you still want an example, try N=2x^2.
Or, try any general value of N like N = P(x) where P(x) is a function in x.

OK. Final argument from my side.
This should be able to stop you.

I assume that you know what integration is and you know how to use it.

So, it's a common thing that if I(D(f(x))) = f(x),
where I(g(x)) means integration of g(x) wrt x.
and, D(h(x)) means integration of h(x) wrt x.

Getting back to the original f(x) = x + x + ..(x times) + x = x^2
So, I say that f'(x) = 2x.
and you say f'(x) = x.

Let's see what we get on integrating :
f(x) = I(x) = (x^2)/2 --- Result according to you.
f(x) = I(2x) = x^2 --- Result according to me.
As per assumption, f(x) = x^2 and not (x^2)/2.
So, your result is clearly WRONG !!

I guess, now you don't have anything more to explain,
neither do I have to clarify anything more.

( Don't tell me you don't know how to integrate ! )

13. ### xpi0t0sMentor

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*** So if you still disagree with this, please provide JUST ONE value of N for which this DOES NOT hold. Should be easy to find. ***

You keep on completely ignoring the above.

Fine, so if I'm completely wrong, which is your argument, this should be REALLY EASY. Never mind the complicated arguments referring to integration, product differentiation, setting f(x)=g(x) and so on. Just ONE explicit value of N please, where f(x)=x+x+x (N times), where x=N, and where f'(x)!=x.

> Getting back to the original f(x) = x + x + ..(x times) + x = x^2

No, this is the problem, you're completely missing the point. f(x)=x+x+x (x times) means that when x=3, f(x)=x+x+x (3 x's - count them). When x=4, f(x)=x+x+x+x. Four x's - count them. x=5: f(x)=x+x+x+x+x. Five x's - count them. Go on, really count them. Tell me there are NOT four x's in "f(x)=x+x+x+x" (on the right hand side of the =).

x+x+x (x times) is phrased that way specifically to mean that I am ***NOT*** talking about f(x)=x^2.

If you don't accept the shorthand f(x)=x+x+x (x times), then use the longhand version instead.
f(x)=f1(x)+f2(x)+f3(x)+...fN(x), where fp(x)=x for p=1..N.
Then what is f(x)?
Now set x=N and tell me what f(x) is.

14. ### xpi0t0sMentor

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> So, I say that f'(x) = 2x.
> and you say f'(x) = x.
> Let's see what we get on integrating :

Yep, OK. I can follow this now. Remember it's a precondition that x=some value N, so let's pick x=N=4.

Int 4 dx = Int 4(x^0) dx = 4x+c. So, not x^2 as you say, but still x+x+x+x which is my original point (plus c).

15. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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You are crazy bro !!

f(x) = x +x + ..(x times) does NOT mean when x = 3, f(x) = 3x.
It means f(x) is variable times the variable.

You are substituting 4 when x times is encountered, but not when 'x' is encountered !!!
You are mistaken there.

OK. Take sqrt(x) instead of 4. What do you do now ??

Another thing,

you can write I(N dx) = NI(dx) = Nx as long as N is a constant.
Otherwise, you can't take N outside the Integration sign.

I found this on googling : http://mathforum.org/library/drmath/view/53415.html

Mistakes happen with everyone, so it's perfectly fine with a Mentor too.
But, don't try to prove that the false thing that you spoke is actually true.

Closing your eyes would only make things worse.

16. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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17. ### xpi0t0sMentor

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OK. I've read that. But it still does not answer the question I have repeatedly asked you and which you apparently cannot answer, or do not want to answer for some reason.

Give me a SINGLE value of N for which my calculation does not apply. If you can do this, then I will accept that I am wrong. But until you can do that then I do not accept that I am wrong. A single counterexample will prove to me that I am wrong and I will happily forfeit the argument.

If there is no value of N for which my calculation does not apply, then my calculation has to be correct.

18. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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OK. Telling it once again won't hurt me.

Try any of the following as your N :
(1) N = sqrt(x),
(2) N = x^2, x^m, p*x^q ...
(3) N = sin x, cos x, log x, exp x ...

For example, let N = sin x
so, f(x) = x sinx
=> f'(x) = sin x as predicted by you.

But even a high school child knows that derivative of x sin x is sin x + x cos x.

Your calculation is incorrect because you are trying to prove that,
there are two different derivatives for x^2, which is simply impossible because it's well-known that,
x^2 is a continuous func with a unique derivative which is 2x.

So, f'(x) = g'(x) = 2x as long as f(x) = g(x) = x^2.

I don't care if you don't accept this. But, fact remains fact .. and I haven't taken the responsibility to enlighten everyone.
As a forum member, it's my duty to guide people in right direction.

But, what can I do, if they want to close their eyes ???

May be you are trying to forcefully prove that the wrong statement that slipped out of your mouth was actually right, just b'coz u are a mentor and that might affect your ego/prestige.

19. ### xpi0t0sMentor

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No you haven't.

> (1) N = sqrt(x),

That is not a VALUE. That is an EXPRESSION, or a FORMULA.

> (2) N = x^2, x^m, p*x^q ...

That is not a VALUE. That is an EXPRESSION, or a FORMULA.

> (3) N = sin x, cos x, log x, exp x ...

That is not a VALUE. That is an EXPRESSION, or a FORMULA.

What I mean by VALUE is a number, say, 2, or 3, or 27. A positive integer. A positive integer value for N for which my expression DOES NOT hold. Because I reckon there isn't one. But since you're convinced I am wrong, then there must be one and it must be easy to find.

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