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.

Answer 1

@SolomonZelman can you help me again?

Answer 2

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

Answer 3

would the derivative of y be 0?

Answer 4

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

Answer 5

ok

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

but you have to substitute back

Answer 11

right

Answer 12

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

Answer 13

so...
\[y"=((3(-2))-((-2(2))(-2(2)/3))/(3^2)\]

Answer 14

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

Answer 15

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

Answer 16

now i say thank you