Question

Calc @zepdrix

Answer 1

hi!!

Answer 2

@zepdrix @Michele_Laino

Answer 3

hi

Answer 4

what is the derivative of \[y=\sqrt{2x}\]?
if you get \[\frac{1}{\sqrt{2x}}\] the answer is YES
if not, NO

Answer 5

Oh, okay! Thank you

Answer 6

is it clear how to do that?

Answer 7

I got \(\Large\frac{\sqrt{2}}{2\sqrt{x}}\) as the derivative, so NO, right?

Answer 8

function \(1/y\) does exist, if ad only if \[\huge x > 0\]

Answer 9

???

Answer 10

if, for example \(x<0\) the function \(y=\sqrt{2x}\) doesn't exist

Answer 11

ah, cause it would be undefined

Answer 12

yes!

Answer 13

\[\large\rm \frac{d}{dx}\sqrt{stuff}=\frac{1}{2\sqrt{stuff}}(stuff)'\]So,\[\large\rm \frac{d}{dx}\sqrt{2x}=\frac{1}{2\sqrt{2x}}(2x)'\]

Answer 14

oh, right. Chain formula right?

Answer 15

Square root derivative, and chain rule, yes.

Answer 16

I think you maybe ended up with the correct derivative,
you just way oversimplified.

Answer 17

I don't really understand why my way doesn't work, because \(\Large\sqrt{2x}=\sqrt{2}x^{1/2}\) right?

Answer 18

Oh I see what you did, hmm

Answer 19

Yes, but that's going to make it more difficult to turn into 1/y is the issue D: hmm

Answer 20

If you do this instead, \(\large\rm \sqrt{2x}=(2x)^{1/2}\)
it's going to work out a lot nicer.

Answer 21

oh, ok. I see

Answer 22

you method works, since we have this correspondence:
\[\huge {x^{1/2}} \to \frac{1}{2}{x^{\frac{1}{2} - 1}}\]

Answer 23

I ended up getting 1/y, so thanks!

Answer 24

:)

Answer 25

yay! \c:/