Question #518f7

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Question #518f7

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Answer 1 · 2018-01-10T04:03:53

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

Explanation:

To find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/dx^2$ , we must first find $\frac{d y}{d x}$ $dy/dx$ which we can do using implicit differentiation.

$\frac{d}{d x} [x^{2} - y^{2}] = \frac{d}{d x} [5]$ $d/dx[x^2-y^2]=d/dx[5]$
$\frac{d}{d x} x^{2} - \frac{d}{d x} y^{2} = \frac{d}{d x} 5$ $d/dxx^2-d/dxy^2=d/dx5$ (Sum Rule)
$2 x - 2 y \frac{d y}{d x} = 0$ $2x-2ydy/dx=0$
$2 x = 2 y \frac{d y}{d x}$ $2x=2ydy/dx$
$\frac{2 x}{2 y} = \frac{d y}{d x}$ $(2x)/(2y)=dy/dx$
$\frac{d y}{d x} = \frac{x}{y}$ $dy/dx=x/y$

Now, we can find the second derivative.

$\frac{d}{d x} [\frac{d y}{d x}] = \frac{d}{d x} [\frac{x}{y}]$ $d/dx[dy/dx]=d/dx[x/y]$
$\frac{d^{2} y}{{d x}^{2}} = \frac{y \frac{d}{d x} x - x \frac{d}{d x} y}{y^{2}}$ $(d^2y)/dx^2=(yd/dxx-xd/dxy)/y^2$ (Quotient Rule)
$\frac{d^{2} y}{{d x}^{2}} = \frac{y - x \frac{d y}{d x}}{y^{2}}$ $(d^2y)/dx^2=(y-xdy/dx)/y^2$
$\frac{d^{2} y}{{d x}^{2}} = \frac{y - x (\frac{x}{y})}{y^{2}}$ $(d^2y)/dx^2=(y-x(x/y))/y^2$
$\frac{d^{2} y}{{d x}^{2}} = \frac{y - \frac{x^{2}}{y}}{y^{2}}$ $(d^2y)/dx^2=(y-x^2/y)/y^2$
$\frac{d^{2} y}{{d x}^{2}} = \frac{y^{2} - x^{2}}{y^{3}}$ $(d^2y)/dx^2=(y^2-x^2)/y^3$

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Question #518f7

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