## asnaseer 3 years ago How good is your understanding of differentiation? $$y=x^2$$ can be written as $$y=x*x$$ which can also be written as $$y=x+x+x+...+x$$ (i.e. "x" lots of x). using this we get:\begin{align} y&=x^2\\ \therefore \frac{dy}{dx}&=2x\\ &\text{but we can also write this as:}\\ y&=x+x+x+...+x\quad\text{i.e. "x" lots of x}\\ \therefore \frac{dy}{dx}&=1+1+1+…+1\quad\text{i.e. "x" lots of 1's}\\ &=x\\ \therefore 2x&=x\\ \therefore 2&=1 \end{align}where is the mistake in this?

1. Mr.Math

In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.

2. asnaseer

it is differentiable and is continuous - just like $$y=x^2$$

3. Mr.Math

Not really. x here can take only integer values, so it's defined from N to R. While y=x^2 is defined from R to R. So they are tow different functions.

4. agdgdgdgwngo

your mistake is that you didn't differentiate the sum of all the x's properly

5. agdgdgdgwngo

but I forgot the exact mistake :-P

6. asnaseer

you can take one and half lots of something

7. Mr.Math

If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?

8. asnaseer

f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

9. TuringTest

how can that be?

10. Mr.Math

Yeah, how can that be?

11. asnaseer

Mr.Math - you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.

12. Akshay_Budhkar

how is f(0.5)= 0.25??

13. asnaseer

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

14. agdgdgdgwngo

or $\sum_{i=1}^{x}x$

15. asnaseer

I didn't what to use that because it implies x must be an integer

16. agdgdgdgwngo

oh

17. ktklown

What does "taken x times" mean then if not a sum?

18. asnaseer

but I guess Mr.Math is right - we cannot say x^2=x+x+x+... (x times) for non-integers?

19. Mr.Math

Just quickly 2x=x means x=0, and doesn't mean 2=1.

20. asnaseer

yes Mr.Math - the real question is, how can I get two different results by differentiating the /same/ function?

21. Akshay_Budhkar

x cannot be zero?

22. asnaseer

at ostensibly the same function

23. asnaseer

*at least

24. agdgdgdgwngo

think of the product rule

25. TuringTest

I don't think you can differentiate a variable number of terms like that, but I can't seem to apply the right rule.

26. Mr.Math

f(x)=x+x+..+x (x times). The key is "x" times, since we're differentiating with respect to x. If it's f(x)=x+x+..+x ("n" times), what we did would be correct.

27. asnaseer

Mr.Math has hit the nail on the head!

28. TuringTest

but which rule exactly are we breaking and how?

29. Akshay_Budhkar

the product rule?

30. Mr.Math

then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.

31. agdgdgdgwngo

right

32. asnaseer

if you re-did this using the definition of differentiation then you would get the right result. i.e.$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

33. asnaseer

@TuringTest - the being broken is to treat "x" as a constant when it is not

34. asnaseer

*the rule

35. Mr.Math

Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.

36. TuringTest

that's what I said, a variable number of terms. I just want to make it concrete in my mind.

37. asnaseer

I guess I've also learnt a thing or do from your replies on what a continuous function is - thx all

38. asnaseer

*or two

39. Akshay_Budhkar

but at the first place is x+x+x+... xtimes a continuous function to be differentiated?

40. asnaseer

@Akshay Budhkar - if you use this to evaluate the differential of y = x+x+...+x (x times), then you will get the right answer:$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

41. agdgdgdgwngo

btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(

42. asnaseer

that highlights the fact that you cannot treat the x in "x times" as a constant

43. asnaseer

@agdgdgdgwngo - thats me in the pic - why? do I look odd?

44. Akshay_Budhkar

No i was just wondering if the function was continuous, i know that formula, but if a function is not continuous we cannot differentiate it, and i feel this is not a continuous function

45. asnaseer

yes - Mr.Math pointed that out earlier - I guess I used a bad example of trying to illustrate a specific point :-(

46. Mr.Math

Wait!! $f'(x)=\lim_{h\to 0}{x+h+x+h+\dots+x+h-x-x-\dots-x \over h}$ $=\lim_{x\to 0}{xh \over h}=x$

47. Mr.Math

$$h\to 0$$*

48. asnaseer

no Mr.Math - the first set of "x+h" terms will appear "x+h" times

49. FoolForMath

This is a old problem.

50. Akshay_Budhkar

you came late ffm that is why it is old :P

51. asnaseer

yes FFM - I came across it several years ago and just recalled it :-)

52. FoolForMath

There has been a discussion on this one, let me do some digging.

53. asnaseer

you mean here on OS?

54. Mr.Math

Right! $\lim_{h \to 0}{x+h+x+h+..+x+h-x-x-\dots -x \over h}$ $=\lim_{h \to 0}{hx+(x+h)h \over h}=2x$

55. asnaseer

thats it Mr.Math - well done!

56. Akshay_Budhkar

So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?

57. asnaseer

the mistake was in treating the "x" in "x times" as a "constant" when differentiating - this is wrong

58. Akshay_Budhkar

yea get it :D

59. FoolForMath
60. asnaseer

I wasn't aware of that FFM - I ran into this many many years ago at uni

61. TuringTest

they mention the integer issue as well...

62. FoolForMath

It's okay, I can't imagine why you mean many many and many :D

63. asnaseer

;-D

64. Akshay_Budhkar

yea @turing