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

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

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Answer 1 · 2016-11-25T07:20:15

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

Explanation:

To find $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ is to find the second differentiation of the
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given expression .
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$5 y^{2} + 2 = 5 x^{2}$ $5y^2+2=5x^2$
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$\frac{d}{d x} (5 y^{2} + 2) = \frac{d}{d x} (5 x^{2})$ $d/dx(5y^2+2)=d/dx(5x^2)$
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$\Rightarrow 10 y (\frac{d y}{d x}) + 0 = 10 x$ $rArr10y(dy/dx)+0=10x$
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$\Rightarrow 10 y (\frac{d y}{d x}) = 10 x$ $rArr10y(dy/dx)=10x$
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$\Rightarrow \frac{d y}{d x} = \frac{10 x}{10 y}$ $rArrdy/dx=(10x)/(10y)$
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Therefore,
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$\frac{d y}{d x} = \frac{x}{y}$ $color(blue)(dy/dx=x/y$
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Let us compute $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/dx^2$
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$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (\frac{d y}{d x})$ $(d^2y)/dx^2=d/dx(dy/dx)$
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$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (\frac{x}{y})$ $(d^2y)/dx^2=d/dx(x/y)$
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Differentiating $\frac{x}{y}$ $x/y$ is determined by applying the quotient rule differentiation.
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$\frac{d^{2} y}{{d x}^{2}} = \frac{(\frac{d x}{d x}) \times y - (\frac{d y}{d x}) \times x}{y^{2}}$ $(d^2y)/dx^2=((dx/dx)xxy-(dy/dx)xx x)/y^2$
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$\frac{d^{2} y}{{d x}^{2}} = \frac{y - x \frac{d y}{d x}}{y^{2}}$ $(d^2y)/dx^2=(y-xcolor(blue)(dy/dx))/y^2$
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$\frac{d^{2} y}{{d x}^{2}} = \frac{y - x (\frac{x}{y})}{y^{2}}$ $(d^2y)/dx^2=(y-x(x/y))/y^2$
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$\frac{d^{2} y}{{d x}^{2}} = \frac{y - \frac{x^{2}}{y}}{y^{2}}$ $(d^2y)/dx^2=(y-x^2/y)/y^2$
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Therefore,
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$\frac{d^{2} y}{{d x}^{2}} = \frac{y^{2} - x^{2}}{y^{3}}$ $(d^2y)/dx^2=(y^2-x^2)/y^3$

Answer 2 · 2016-11-25T07:33:30

The answer is $= - \frac{2}{5} {(\frac{5 x^{2} - 2}{5})}^{- \frac{3}{2}}$ $=-2/5((5x^2-2)/5)^(-3/2)$

Explanation:

Let's do the implicit differentiation

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

$10 y \frac{d y}{d x} + 0 = 10 x$ $10ydy/dx+0=10x$

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

This is of the form $\frac{u}{v}$ $u/v$

$(u/v)'=(u'v-uv')/v^2$

So, $(d^2y)/dx^2=(x/y)'=(y-xdy/dx)/y^2$

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

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

$y^2=(5x^2-2)/5$

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

Therefore, $(d^2y)/dx^2=((5x^2-2)/5-x^2)/((5x^2-2)/5)^(3/2)$

$=-2/(5((5x^2-2)/5)^(3/2))$

$=-2/5((5x^2-2)/5)^(-3/2)$

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

2 Answers

Explanation:

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