Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
asnaseer
Group Title
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?
 2 years ago
 2 years ago
asnaseer Group Title
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?
 2 years ago
 2 years ago

This Question is Closed

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
it is differentiable and is continuous  just like \(y=x^2\)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
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.
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
your mistake is that you didn't differentiate the sum of all the x's properly
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
but I forgot the exact mistake :P
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
you can take one and half lots of something
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
how can that be?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
Yeah, how can that be?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
Mr.Math  you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
how is f(0.5)= 0.25??
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
f(x)=x^2=x+x+... (taken x times)
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
or \[\sum_{i=1}^{x}x\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
I didn't what to use that because it implies x must be an integer
 2 years ago

ktklown Group TitleBest ResponseYou've already chosen the best response.0
What does "taken x times" mean then if not a sum?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
but I guess Mr.Math is right  we cannot say x^2=x+x+x+... (x times) for nonintegers?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
Just quickly 2x=x means x=0, and doesn't mean 2=1.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
yes Mr.Math  the real question is, how can I get two different results by differentiating the /same/ function?
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
x cannot be zero?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
at ostensibly the same function
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
think of the product rule
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I don't think you can differentiate a variable number of terms like that, but I can't seem to apply the right rule.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
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.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
Mr.Math has hit the nail on the head!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
but which rule exactly are we breaking and how?
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
the product rule?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
if you redid this using the definition of differentiation then you would get the right result. i.e.\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)f(x)}{h}\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
@TuringTest  the being broken is to treat "x" as a constant when it is not
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
that's what I said, a variable number of terms. I just want to make it concrete in my mind.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
I guess I've also learnt a thing or do from your replies on what a continuous function is  thx all
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
but at the first place is x+x+x+... xtimes a continuous function to be differentiated?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
@Akshay Budhkar  if you use this to evaluate the differential of y = x+x+...+x (x times), then you will get the right answer:\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)f(x)}{h}\]
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
that highlights the fact that you cannot treat the x in "x times" as a constant
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
@agdgdgdgwngo  thats me in the pic  why? do I look odd?
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
yes  Mr.Math pointed that out earlier  I guess I used a bad example of trying to illustrate a specific point :(
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
Wait!! \[f'(x)=\lim_{h\to 0}{x+h+x+h+\dots+x+hxx\dotsx \over h}\] \[=\lim_{x\to 0}{xh \over h}=x\]
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
\(h\to 0\)*
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
no Mr.Math  the first set of "x+h" terms will appear "x+h" times
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
This is a old problem.
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
you came late ffm that is why it is old :P
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
yes FFM  I came across it several years ago and just recalled it :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
There has been a discussion on this one, let me do some digging.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
you mean here on OS?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.5
Right! \[\lim_{h \to 0}{x+h+x+h+..+x+hxx\dots x \over h}\] \[=\lim_{h \to 0}{hx+(x+h)h \over h}=2x\]
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
thats it Mr.Math  well done!
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
the mistake was in treating the "x" in "x times" as a "constant" when differentiating  this is wrong
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
yea get it :D
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Here is it http://math.stackexchange.com/questions/1096/
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.6
I wasn't aware of that FFM  I ran into this many many years ago at uni
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
they mention the integer issue as well...
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
It's okay, I can't imagine why you mean many many and many :D
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
yea @turing
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.