Question

how do you find the the second dy/dx of x^2+y^2 = 4

Answer 1

derivative wrt x right? did you find
\[y'=-\frac{x}{y}\]?

Answer 2

\[2x+2yy'=0 => x+yy'=0 =>yy'=-x= > y'=\frac{-x}{y}\]

Answer 3

because that is the first step if you want
\[y''\] if not let me know and we can find it quickly. but once you have that, it is the quotient rule

Answer 4

ok i really meant quickly clearly!

Answer 5

\[\implies\]

Answer 6

\[y''=\frac{-1y-(-x)y'}{y^2}=\frac{-y+xy'}{y^2}=\frac{-y+x \cdot \frac{-x}{y}}{y^2}\]

Answer 7

\implies

Answer 8

\[=\frac{-y^2-x^2}{y^3}\]

Answer 9

\[=\frac{-y+\frac{-x^2}{y}}{y^2}=\frac{y}{y} \frac{-y +\frac{-x^2}{y}}{y^2} =\frac{-y^2-x^2}{y^3}\]
\[=\frac{-(x^2+y^2)}{y^3}=\frac{-4}{y^3}\]

Answer 10

can someone help me

Answer 11

where did the -4 come from?

Answer 12

don't bite!

Answer 13

\[x^2+y^2=4 => -(x^2+y^2)=-4\]

Answer 14

you bit

Answer 15

\implies

Answer 16

\[x^2+y^2=4\implies -(x^2+y^2)=-4\]

Answer 17

how come my latex isn't showing up nice and pretty?

Answer 18

looks good to me

Answer 19

but my arrows are better

Answer 20

shhh..
i didn't bite hard did i ?

Answer 21

\[x^2+y^2=4\iff -(x^2+y^2)=-4\]

Answer 22

i meant "don't bite" as in "i am baiting you"

Answer 23

lol ok

Answer 24

holdon for the first dy/dx i got -2x/2y and that is right

Answer 25

yes you cancel the two's though

Answer 26

ohh lol alright i see ty