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asnaseer
 5 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
 5 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?

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Mr.Math
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.6it is differentiable and is continuous  just like \(y=x^2\)

Mr.Math
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0but I forgot the exact mistake :P

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

Mr.Math
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.6f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

asnaseer
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0how is f(0.5)= 0.25??

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

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

asnaseer
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0What does "taken x times" mean then if not a sum?

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

asnaseer
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0x cannot be zero?

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

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

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

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

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

Mr.Math
 5 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
 5 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
 5 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
 5 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
 5 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
 5 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
 5 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
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.6@agdgdgdgwngo  thats me in the pic  why? do I look odd?

Akshay_Budhkar
 5 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
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0This is a old problem.

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

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

anonymous
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.6thats it Mr.Math  well done!

Akshay_Budhkar
 5 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
 5 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
 5 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
 5 years ago
Best ResponseYou've already chosen the best response.0they mention the integer issue as well...

anonymous
 5 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
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