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unimatix
 one year ago
For delta x infinitesimal find an infinitesimal (epsilon) such that delta y = dy + epsilon delta x
2sqrt(x)
unimatix
 one year ago
For delta x infinitesimal find an infinitesimal (epsilon) such that delta y = dy + epsilon delta x 2sqrt(x)

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unimatix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441743136199:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441743200586:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you please use the equation button for this I do not understand the writing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks like the definition of a derivative

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Delta y = dy + \epsilon \Delta x\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0And I think I've found dy and delta y. Which are \[\Delta y = \frac{ 2 \Delta x }{ \sqrt{x+ \Delta x} + \sqrt{x} }\] \[dy = \frac{ \Delta x }{ \sqrt{x} }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh... this relates to the increment theorem... vaguely remember learning about this x_x

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Yep, that sounds right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im just giving a shot at this,...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(y=2\sqrt{x}\) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we're ultimately trying to put it in the form of \[\frac{\Delta y}{\Delta x} = f'(x) + \epsilon\] right?...

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Delta y = dy+ \epsilon \Delta x\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I've just got to find epsilon.

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0\[\Delta y = \frac{ 1 }{ 2\sqrt{x} } \Delta x + \epsilon \Delta x\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \Delta y }{ \Delta x } = \frac{ 1 }{ 2\sqrt{x}} + \epsilon \]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0\[\epsilon = \frac{ \Delta y }{ \Delta x }  \frac{ 1 }{ 2\sqrt{x} }\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Maybe substitute back in the Delta y from earlier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh instead of finding "\(\epsilon\)" can't we take the limit? Just throwing that out there, i might be wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok for y = dy + epsilon delta x what does this even mean :O

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I'm sure there is a perfectly good explanation for that...

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0@Halmos Maybe it's meant to be? dw:1441745412913:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=2\sqrt{x}\]\[y+\Delta y = 2\sqrt{x+\Delta x}\]\[y+\Delta y y = 2\sqrt{x+\Delta x}2\sqrt{x}\]\[\Delta y =2\sqrt{x+\Delta x}2\sqrt{x}\]hmm.....

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0IDK, @unimatix you need to check the problem statement. Either that or your text uses some very nonstandard terminology.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you wanna compare this equation to Increment Epsilon Form then you would got \(\epsilon =\dfrac{y}{\Delta x}  \dfrac{dy}{\Delta x} \\\epsilon =\dfrac{dy}{dx}\dfrac{\Delta y}{\Delta x} \) ermmm first ewuation which is in ur question it does not make sense how would you interpret it ?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0@beginnersmind I double checked the problem. I'm using https://www.math.wisc.edu/~keisler/calc.html section 2.2 Lots of people seem to like the book online so idk. It is a bit old.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\Delta y =\frac {2\sqrt{x+\Delta x}2\sqrt{x}}{1} \cdot \frac{2\sqrt{x+\Delta x}+2\sqrt{x}}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4(x+\Delta x)4x}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4\Delta x}{2\sqrt{x+\Delta x} +2\sqrt{x}}\]\[\frac{\Delta y}{\Delta x}= \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which problem is it ?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Oh, right, I think I read parts of this book a few years ago. It definitely uses an unusual approach :) "This is a calculus textbook at the college Freshman level based on Abraham Robinson's infinitesimals, which date from 1960. Robinson's modern infinitesimal approach puts the intuitive ideas of the founders of the calculus on a mathematically sound footing, and is easier for beginners to understand than the more common approach via limits. "

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0It introduces definitions for dy and dx at the start so you can do algebra with them, beyond the kind of chain rule cancellation you do in "normal" calculus classes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Taking the limit of the function...\[\lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x} = \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}} = \frac{4}{4\sqrt{x}} =\frac{1}{\sqrt{x}}\]So...... \[\frac{dy}{dx} = \frac{1}{\sqrt{x}}\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Note the title of chapter 1: "Real and Hyperreal numbers"

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I didn't find Chapter 1 that bad.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, but how do we go about solving the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\Delta x \approx dx \) and we already know \(dy=f'(x) dx\) Increment theorem would be this cuz of the assumption \(\Delta x = dx \) \( \Delta y =dy+\epsilon dx \) that make since now :D

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0@unimatrix It's not that it's bad. It's just that if someone didn't read the book they literally won't know what dz and dx means in this problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dz is using partials?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Chapter 1 defines them in a specific way, that's slightly different from what you'd see in a normal calculus book or course.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I love how they use \(d\) for derivative. That makes me happy.
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