Question

need help on the attachement

Answer 1

that is the second derivitive

Answer 2

take the first derivitve.. 2x+4y
second 2+4 = 6

Answer 3

does that make sense ta?

Answer 4

did you start with
\[2x+4yy'=0\]
\[y'=-\frac{x}{2y}\]

Answer 5

yes!

Answer 6

no it doesn't make sense. you are looking for the second derivative wrt x

Answer 7

yeah, i was wrong.. thanks satellite..

Answer 8

ok now we take the derivative again using the quotient rule.

Answer 9

I got 2y^(2)+x^(2)/2y^(3)

Answer 10

when taking the second derivative

Answer 11

let me write it with paper so i don't screw up

Answer 12

ok

Answer 13

ok a first step is
\[-\frac{1}{2}\frac{y-xy'}{y^2}\]

Answer 14

so replace
\[y'=-\frac{x}{2y}\] and i will do the algebra on paper as well

Answer 15

if i didn't mess up the algebra i get
\[-\frac{1}{2}\frac{2y^2+x^2}{2y^3}\]

Answer 16

or if you prefer
\[-\frac{2y^2+x^2}{4y^2}\]

Answer 17

looks like you got the same thing exactly but maybe forgot the
\[-\frac{1}{2}\] out front when you simplified it

Answer 18

Oops! Can't forget negative, lol. However on the answer choices have -1/2y^(2) why?

Answer 19

Also why is the top of numerator still 2y^(2)

Answer 20

i don't know.

Answer 21

let me write it out

Answer 22

we will start with
\[-\frac{1}{2}\frac{y-xy'}{y^2}\] ok?

Answer 23

then
\[-\frac{1}{2}\frac{y-x\frac{-x}{2y}}{y^2}\]
\[-\frac{1}{2}\frac{y+x\frac{x}{2y}}{y^2}\]

Answer 24

ok

Answer 25

then multiply numerator and denominator by
\[2y\] to clear the fraction and get
\[-\frac{1}{2}\frac{2y^2+x^2}{2y^3}\]

Answer 26

and then perhaps
\[-\frac{2y^2+x^2}{4y^3}\]

Answer 27

but can't you cancel 2 and the 4

Answer 28

it is always possible that i made an algebra mistake, but it looks good to me

Answer 29

here the answer choices

Answer 30

oh i see what you are asking. no you cannot for the same reason that
\[\frac{2\times 5+7}{4\times 3}\neq \frac{5+7}{2\times 3}\]

Answer 31

lol i wasn't thinking. ok look at the original equation, and our answer

Answer 32

here is the answer
\[-\frac{2y^2+x^2}{4y^3}\] right? but the original equation was
\[x^2+2y^2=2\] so our numerator is a big fat 2

Answer 33

very tricky...

Answer 34

so we get
\[-\frac{2}{4y^3}=-\frac{1}{2y^3}\]

Answer 35

damn i still don't see that answer on your choices. am i missing one?

Answer 36

yeah this was actually the first answer choice

Answer 37

why would they do that trick?

Answer 38

to get on your last nerve

Answer 39

its should be the previous answer, lol