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?

Question

mr.math · Answer

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

asnaseer · Answer

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

mr.math · Answer

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.

anonymous · Answer

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

anonymous · Answer

but I forgot the exact mistake :-P

asnaseer · Answer

you can take one and half lots of something

mr.math · Answer

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

asnaseer · Answer

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

turingtest · Answer

how can that be?

mr.math · Answer

Yeah, how can that be?

asnaseer · Answer

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

akshay_budhkar · Answer

how is f(0.5)= 0.25??

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

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

anonymous · Answer

oh

anonymous · Answer

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

asnaseer · Answer

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

mr.math · Answer

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

asnaseer · Answer

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

akshay_budhkar · Answer

x cannot be zero?

asnaseer · Answer

at ostensibly the same function

asnaseer · Answer

*at least

anonymous · Answer

think of the product rule

turingtest · Answer

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

mr.math · Answer

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.

asnaseer · Answer

Mr.Math has hit the nail on the head!

turingtest · Answer

but which rule exactly are we breaking and how?

akshay_budhkar · Answer

the product rule?

mr.math · Answer

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

anonymous · Answer

right

asnaseer · Answer

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

asnaseer · Answer

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

asnaseer · Answer

*the rule

mr.math · Answer

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

turingtest · Answer

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

asnaseer · Answer

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

asnaseer · Answer

*or two

akshay_budhkar · Answer

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

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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

akshay_budhkar · Answer

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

asnaseer · Answer

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

mr.math · Answer

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

mr.math · Answer

$h\to 0$*

asnaseer · Answer

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

anonymous · Answer

This is a old problem.

akshay_budhkar · Answer

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

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

you mean here on OS?

mr.math · Answer

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

asnaseer · Answer

thats it Mr.Math - well done!

akshay_budhkar · Answer

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

asnaseer · Answer

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

akshay_budhkar · Answer

yea get it :D

anonymous · Answer

Here is it http://math.stackexchange.com/questions/1096/

asnaseer · Answer

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

turingtest · Answer

they mention the integer issue as well...

anonymous · Answer

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

asnaseer · Answer

;-D

akshay_budhkar · Answer

yea @turing