If 2x^2+y^2=-2 then evaluate d^2y/dx^2 when x=2 and y=3. Round your answer to 2 decimal places.

Question

If 2x^2+y^2=-2 then evaluate d^2y/dx^2 when x=2 and y=3.  Round your answer to 2 decimal places.

babbs12 · Answer

@SolomonZelman can you help me again?

owen3 · Answer

2x^2+y^2=-2 

2(2x) + 2y * dy/dx = 0
4x + 2y dy/dx = 0

its easier if we use y ' to stand for dy/dx

4x + 2y * y ' = 0
y' = -4x /(2y) 
y' = -2x / y 

now take derivative again

babbs12 · Answer

would the derivative of y be 0?

owen3 · Answer

nope, derivative of y is dy/dx,  or y'

babbs12 · Answer

ok

babbs12 · Answer

so it would be 
(y'*-2x)-(-2*y)y^2
right?

babbs12 · Answer

(y'*-2x)-(-2*y)/y^2 is what i meant

owen3 · Answer

y' = -2x / y 
y ' ' = ( y(-2)  - (-2x)*y ' ) / y^2

babbs12 · Answer

thank you...wasn't sure where to start on it

owen3 · Answer

but you have to substitute back

babbs12 · Answer

right

owen3 · Answer

this is what you should get http://prntscr.com/9puxo8

babbs12 · Answer

so...
$$y"=((3(-2))-((-2(2))(-2(2)/3))/(3^2)$$

owen3 · Answer

you should get -34/27 http://prntscr.com/9puynv

babbs12 · Answer

y"=((-6)-(-4*(-4/3)))/9
=(-6-16/3)/9
=(-18/3-16/3)/9
=(-34/3)/9
=-34/27

babbs12 · Answer

now i say thank you