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

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Answer 1 · 2016-10-14T18:44:59

Deleted, because it was incorrect

Answer 2 · 2016-10-14T19:43:40

I get $-5/(4y^3)$

$2y^2=x^2-5$

$4y dy/dx = 2x$

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

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

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

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

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

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

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

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

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

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

Answer 3 · 2016-10-14T22:47:23

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

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

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

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

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

How do you find (d^2y)/(dx^2)(d^2y)/(dx^2) for 5=x^2-2y^25=x^2-2y^2?