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

Nov 25, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{y}^{2} - {x}^{2}}{y} ^ 3$

#### Explanation:

To find $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ is to find the second differentiation of the
$\text{ }$
given expression .
$\text{ }$
$5 {y}^{2} + 2 = 5 {x}^{2}$
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$\frac{d}{\mathrm{dx}} \left(5 {y}^{2} + 2\right) = \frac{d}{\mathrm{dx}} \left(5 {x}^{2}\right)$
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$\Rightarrow 10 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 0 = 10 x$
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$\Rightarrow 10 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 10 x$
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$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 x}{10 y}$
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Therefore,
$\text{ }$
color(blue)(dy/dx=x/y
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Let us compute $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{x}{y}\right)$
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Differentiating $\frac{x}{y}$ is determined by applying the quotient rule differentiation.
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\left(\frac{\mathrm{dx}}{\mathrm{dx}}\right) \times y - \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \times x}{y} ^ 2$
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{y - x \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}}}}{y} ^ 2$
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{y - x \left(\frac{x}{y}\right)}{y} ^ 2$
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{y - {x}^{2} / y}{y} ^ 2$
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Therefore,
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$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{{y}^{2} - {x}^{2}}{y} ^ 3$

Nov 25, 2016

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

#### Explanation:

Let's do the implicit differentiation

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

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} + 0 = 10 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$

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

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

So, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(\frac{x}{y}\right) ' = \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2$

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

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

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

${y}^{3} = {\left(\frac{5 {x}^{2} - 2}{5}\right)}^{\frac{3}{2}}$

Therefore, $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\frac{5 {x}^{2} - 2}{5} - {x}^{2}}{\frac{5 {x}^{2} - 2}{5}} ^ \left(\frac{3}{2}\right)$

$= - \frac{2}{5 {\left(\frac{5 {x}^{2} - 2}{5}\right)}^{\frac{3}{2}}}$

$= - \frac{2}{5} {\left(\frac{5 {x}^{2} - 2}{5}\right)}^{- \frac{3}{2}}$