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

- unimatix

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- unimatix

|dw:1441743136199:dw|

- unimatix

|dw:1441743200586:dw|

- anonymous

Can you please use the equation button for this I do not understand the writing.

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## More answers

- unimatix

sure

- Jhannybean

Looks like the definition of a derivative

- unimatix

\[\Delta y = dy + \epsilon \Delta x\]

- unimatix

And I think I've found dy and delta y. Which are
\[\Delta y = \frac{ 2 \Delta x }{ \sqrt{x+ \Delta x} + \sqrt{x} }\]
\[dy = \frac{ \Delta x }{ \sqrt{x} }\]

- Jhannybean

Ahh... this relates to the increment theorem... vaguely remember learning about this x_x

- unimatix

Yep, that sounds right

- Jhannybean

Im just giving a shot at this,...

- Jhannybean

\(y=2\sqrt{x}\) right?

- unimatix

Yep!

- Jhannybean

whew, ok.

- Jhannybean

we're ultimately trying to put it in the form of \[\frac{\Delta y}{\Delta x} = f'(x) + \epsilon\] right?...

- unimatix

\[\Delta y = dy+ \epsilon \Delta x\]

- unimatix

I've just got to find epsilon.

- unimatix

\[\Delta y = \frac{ 1 }{ 2\sqrt{x} } \Delta x + \epsilon \Delta x\]

- unimatix

\[\frac{ \Delta y }{ \Delta x } = \frac{ 1 }{ 2\sqrt{x}} + \epsilon \]

- unimatix

\[\epsilon = \frac{ \Delta y }{ \Delta x } - \frac{ 1 }{ 2\sqrt{x} }\]

- unimatix

Maybe substitute back in the Delta y from earlier.

- Jhannybean

Oh instead of finding "\(\epsilon\)" can't we take the limit? Just throwing that out there, i might be wrong

- anonymous

ok for
y = dy + epsilon delta x
what does this even mean :O

- unimatix

I'm sure there is a perfectly good explanation for that...

- beginnersmind

@Halmos Maybe it's meant to be?
|dw:1441745412913:dw|

- Jhannybean

\[y=2\sqrt{x}\]\[y+\Delta y = 2\sqrt{x+\Delta x}\]\[y+\Delta y -y = 2\sqrt{x+\Delta x}-2\sqrt{x}\]\[\Delta y =2\sqrt{x+\Delta x}-2\sqrt{x}\]hmm.....

- beginnersmind

IDK, @unimatix you need to check the problem statement. Either that or your text uses some very non-standard terminology.

- anonymous

if 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

@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.

- Jhannybean

\[\Delta y =\frac {2\sqrt{x+\Delta x}-2\sqrt{x}}{1} \cdot \frac{2\sqrt{x+\Delta x}+2\sqrt{x}}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4(x+\Delta x)-4x}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]\[\Delta y = \frac{4\Delta x}{2\sqrt{x+\Delta x} +2\sqrt{x}}\]\[\frac{\Delta y}{\Delta x}= \frac{4}{2\sqrt{x+\Delta x}+2\sqrt{x}}\]

- anonymous

which problem is it ?

- anonymous

http://prntscr.com/8e5m1s

- anonymous

-..-

- beginnersmind

Oh, 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

It 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.

- Jhannybean

Taking 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

Note the title of chapter 1: "Real and Hyperreal numbers"

- unimatix

I didn't find Chapter 1 that bad.

- Jhannybean

Okay, but how do we go about solving the problem?

- anonymous

\(\Delta x \approx dx \) 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

@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.

- Jhannybean

dz is using partials?

- beginnersmind

Chapter 1 defines them in a specific way, that's slightly different from what you'd see in a normal calculus book or course.

- Jhannybean

I love how they use \(d\) for derivative. That makes me happy.

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