# How do you find (d^2y)/(dx^2) for 5x^2=5y^2+4?

Nov 11, 2016

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

#### Explanation:

The equation does not express $y$ explicitly in terms of $x$ as in $y = f \left(x\right)$, instead we have $g \left(y\right) = f \left(x\right)$, so when we differentiate we apply the chain rule so that we differentiate $g \left(y\right)$ wrt $y$ rather than $x$, as in:

$g \left(y\right) = f \left(x\right)$
$\therefore \frac{d}{\mathrm{dx}} g \left(y\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} g \left(y\right) = f ' \left(x\right)$
$\therefore g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

So for $5 {x}^{2} = 5 {y}^{2} + 4$ we have:

$10 x = 10 y \frac{\mathrm{dy}}{\mathrm{dx}}$
$y \frac{\mathrm{dy}}{\mathrm{dx}} = x$ ..... [1]

To find the second derivative we need to differentiate a second time again implicitly and using the product rule:

$y \frac{\mathrm{dy}}{\mathrm{dx}} = x$
$\therefore \left(y\right) \left(\frac{d}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(\frac{d}{\mathrm{dx}} y\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1$
$\therefore y \left({d}^{2 y} / \left({\mathrm{dx}}^{2}\right)\right) + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 1$

From [1] we have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$ so substituting gives:

$\therefore y \left({d}^{2 y} / \left({\mathrm{dx}}^{2}\right)\right) + {\left(\frac{x}{y}\right)}^{2} = 1$
$\therefore y {d}^{2 y} / \left({\mathrm{dx}}^{2}\right) = 1 - {x}^{2} / {y}^{2}$
$\therefore y {d}^{2 y} / \left({\mathrm{dx}}^{2}\right) = \frac{{y}^{2} - {x}^{2}}{y} ^ 2$

so we have

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

But from the original equation we have;
$5 {x}^{2} = 5 {y}^{2} + 4 \implies 5 \left({y}^{2} - {x}^{2}\right) = - 4 \implies {y}^{2} - {x}^{2} = - \frac{4}{5}$

Hence,

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