calc,

https://www.dropbox.com/s/o2v0t0mlnzr4djz/Screenshot%202015-12-07%2019.28.49.png?dl=0

Question

calc,


https://www.dropbox.com/s/o2v0t0mlnzr4djz/Screenshot%202015-12-07%2019.28.49.png?dl=0

hartnn · Answer

you got dy/dx first?

thadds2003 · Answer

so you would have to combine like terms......

hartnn · Answer

i'll skip some steps,

$2x^2 + 2y^2 y' = 0 \ x^2 + y^2 y' =0 ...(A)\ y' = -x^2/y^2 ....(B) \ From (A), 2x+ 2yy' + y^2 y'' = 0 \ 2x+2y(-x^2/y^2) + y^2 y'' = 0 $

isolate y''

christos · Answer

can I just find dy/dx twice ?

christos · Answer

dy/dx = -x^2/y^2

christos · Answer

now (-x^2/y^2)' to find d^2y/dx^2

hartnn · Answer

ofcourse.

you need to use quotient rule then.

I try to avoid that as much as possible, and use product rule wherever i can.

christos · Answer

Ok because I dont understand the first method ..

hartnn · Answer

go with quotient rule then :)
we'll match our answers to verify!

christos · Answer

ok!

hartnn · Answer

let me know what u get, when u get...or ask if you get stuck...

you'll have to use dy/dx = -x^2/y^2 again/

christos · Answer

i got y/x

hartnn · Answer

|dw:1449510476763:dw|
got that first step? ^^