Question

Find the second derivative of x^2-xy+y^2=1
I keep coming up with -6x^2+6xy-6y^2/(2y-x)^3
but the answer is -6/(2y-x)^3

Answer 1

what did you get for the first derivative?

Answer 2

\[\frac{ y-2x }{ 2y-x }\]

Answer 3

which is correct

Answer 4

ok that looks good

Answer 5

so taking the second derivative, you are going to get an expression that contains the first derivative right?

Answer 6

yep

Answer 7

i used quotient rule

Answer 8

my guess is, without doing the problem, that when you see \(y'\) in the second derivative, replace it by \(\frac{2-2x}{2y-x}\)
also you may get somewhere \(x^2-xy+y^2\) which you can replace by 1
in fact i think i see it in your answer

Answer 9

\[\frac{ (2y-x)(y'-2)-(y-2x)(2y'-1) }{ (2y-x)^2 }\]

Answer 10

hold on
your answer is right i can see it

Answer 11

factor a 6 out of it

Answer 12

i have a +6xy though

Answer 13

\[ -6x^2+6xy-6y^2\]is your numerator yes?

Answer 14

yes

Answer 15

factor out a \(-6\) get
\[-6(x^2-2y+y^2)\]

Answer 16

which is \(-6\) in your second answer as \(x^2-2xy+y^2=1\)

Answer 17

its not 2 xy though

Answer 18

yeah i made a typo is all

Answer 19

it becomes
-6(x^2-xy+y^2)

Answer 20

right
which is -6

Answer 21

does that acually turn in to 1?

Answer 22

you started with
\[x^2-xy+y^2=1\] right?

Answer 23

oh i see it

Answer 24

whew

Answer 25

that makes sense

Answer 26

sorry about the typos

Answer 27

its all fine