A community for students.
Here's the question you clicked on:
 0 viewing
asnaseer
 4 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?
asnaseer
 4 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?

This Question is Closed

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

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

Mr.Math
 4 years ago
Best ResponseYou've already chosen the best response.5Not 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
 4 years ago
Best ResponseYou've already chosen the best response.0your mistake is that you didn't differentiate the sum of all the x's properly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but I forgot the exact mistake :P

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

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or \[\sum_{i=1}^{x}x\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What does "taken x times" mean then if not a sum?

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

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

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

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.0x cannot be zero?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.6at ostensibly the same function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0think of the product rule

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.5f(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
 4 years ago
Best ResponseYou've already chosen the best response.6Mr.Math has hit the nail on the head!

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

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.0the product rule?

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

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

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

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

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

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

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

asnaseer
 4 years ago
Best 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:\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)f(x)}{h}\]

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

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

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

Akshay_Budhkar
 4 years ago
Best ResponseYou've already chosen the best response.0No 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
 4 years ago
Best ResponseYou've already chosen the best response.6yes  Mr.Math pointed that out earlier  I guess I used a bad example of trying to illustrate a specific point :(

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is a old problem.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There has been a discussion on this one, let me do some digging.

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

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

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

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0they mention the integer issue as well...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's okay, I can't imagine why you mean many many and many :D
Ask your own question
Sign UpFind more explanations on OpenStudy
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.