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

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

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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 $- \frac{5}{4 y^{3}}$ $-5/(4y^3)$

Explanation:

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

$4 y \frac{d y}{d x} = 2 x$ $4y dy/dx = 2x$

$\frac{d y}{d x} = \frac{x}{2 y}$ $dy/dx = x/(2y)$

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{x}{2 y})$ $d/dx(dy/dx) = d/dx(x/(2y))$

$= \frac{(1) (2 y) - x (2 (\frac{d y}{d x}))}{{(2 y)}^{2}}$ $= ((1)(2y)-x(2(dy/dx)))/(2y)^2$

$= \frac{2 y - 2 x (\frac{x}{2 y})}{4 y^{2}}$ $= (2y-2x(x/(2y)))/(4y^2)$

$= \frac{y - x (\frac{x}{2 y})}{2 y^{2}}$ $= (y-x(x/(2y)))/(2y^2)$

$= \frac{y - x (\frac{x}{2 y})}{2 y^{2}} \cdot \frac{2 y}{2 y}$ $= (y-x(x/(2y)))/(2y^2) * (2y)/(2y)$

$= \frac{2 y^{2} - x^{2}}{4 y^{3}}$ $= (2y^2-x^2)/(4y^3$

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

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

$\frac{d^{2} y}{{d x}^{2}} = - \frac{5}{4 y^{3}}$ $(d^2y)/dx^2 = -5/(4y^3)$

Answer 3 · 2016-10-14T22:47:23

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

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

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

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

3 Answers

Explanation:

Explanation:

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