Question

implicit differentiation

http://gyazo.com/8e1b2014bebfd49308474182b88751ce

Answer 1

what u got dy/dx as ?

Answer 2

lost connection. sec

Answer 3

dy/dx=4x^3/-4y^3

Answer 4

which is -x^3/y^3 , right ?
but i would suggest u to keep that as
y^3 y' = -x^3
now diff again w.r.t x

Answer 5

use product rue for y^3y'

Answer 6

you lost me there. i differntiated once. why am i doing it again? :|

Answer 7

\(\huge \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx}=y''\)

Answer 8

and it should be x^3/-y^3

Answer 9

but u got dy/dx = -x^3/y^3 =y' , right ?

Answer 10

so im basically taking the 2nd deriviative? and sorry keep losing connection

Answer 11

yes!
d^2y / dx^2 = 2nd derivative

Answer 12

i got dy/dx = x^3/-y^3

Answer 13

which is -x^3/y^3 , right ?
but i would suggest u to keep that as
y^3 y' = -x^3
now diff again w.r.t x

Answer 14

unless x^3/-y^3 is the same thing as what your writing

Answer 15

how do i go about taking diff with respect to x :S

Answer 16

the same exact way, u took first derivative and found dy/dx
u again take derivative w.r.t x
d/dx(dy/dx) = d^2y/dx^2 = y''

Answer 17

so i take derivative of -x^3/y^3?

Answer 18

-3x^2/3y^2?

Answer 19

you can, u need to use quotient rule then, but why use quotion rule when u can use product rule in y^3 y' = -x^3
and no, thats incorrect.

Answer 20

remember 'y' is the function of x
so you cannot write d/dx (y^3) as 3y^2
and u haven't heard of quotient rule ?

Answer 21

ok. let me get this straight. first, i sovled for dy/dx for the original equation and got x^3/-y^3, which is same thing as y^3 y' = -x^3. now, i take the derivivative once again, with respect to x using product rule.

Answer 22

u have 2 options :
y' = - x^3 / y^3 , diff. again w.r.t x using quotient rule
or
y^3 y' = -x^3. , diff.again w..r.t x using product rule
suit yourself
2nd option will be less complicated

Answer 23

ok. so lets do product rule. (3y^2)(y'')=-3x^2?

Answer 24

i messed that up. sec.

Answer 25

(3y^2)(y')+(y^3)(y'')=-3x^2

not sure what to do with y'

Answer 26

now, put y' = - x^3 / y^3
and don't simplify
then put x=0, y=4
and u get your y'' , easily

Answer 27

whoa nelly. ok let me see. now what about y''

Answer 28

got (3y^2)(-x^3/y^3)+(y^3)(y'')=-3x^2 is what i have so far

Answer 29

right, nw just put x=0, y= 4
the entire 1st term = 0

Answer 30

the whole thing will be zero since there are x's on both sides

Answer 31

u get 4^3 y'' =0
---> y''=0
that would be your answer

Answer 32

all that work just to have my answer be zero. :<

Answer 33

what ever happened to the y'' exactly. should that be some sort of function as well?

Answer 34

u can find y'' from here (3y^2)(-x^3/y^3)+(y^3)(y'')=-3x^2
by isolating y''
but since we wanted y'' at a point, we can just put that point in this equation also

Answer 35

ahh ic. well thanks for your help. you just helped me for the past 30 minutes on a friday night.

Answer 36

welcome ^_^
its sat. morning here

Answer 37

o ye. saturday morning here too now lol.