How do you find (d^2y)/(dx^2) for 5=x^2-2y^2?

3 Answers
Oct 14, 2016

Deleted, because it was incorrect

Oct 14, 2016

I get -5/(4y^3)

Explanation:

2y^2=x^2-5

4y dy/dx = 2x

dy/dx = x/(2y)

d/dx(dy/dx) = d/dx(x/(2y))

= ((1)(2y)-x(2(dy/dx)))/(2y)^2

= (2y-2x(x/(2y)))/(4y^2)

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

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

= (2y^2-x^2)/(4y^3

We started with 5=x^2-2y^2, so we have

2y^2-x^2=-5, making the second derivative,

(d^2y)/dx^2 = -5/(4y^3)

Oct 14, 2016

y'' = -5/4 1/y^3

Explanation:

2y(x)^2-x^2+5=0->d/(dx)(2y(x)^2-x^2+5)=4y y'-2x=0

d/(dx)(2y y'-x)=2((y')^2+y y'')-1=0 so

y''=(1/2-(y')^2)/y but y'=1/2x/y so

y'' = (2y^2-x^2)/(4y^3) = -5/4 1/y^3