Feb 23, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{8}{y} - \frac{64 {x}^{2}}{y} ^ 3$

Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

$8 {x}^{2} + {y}^{2} = 2$

Differentiate wrt $x$:

$16 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 8 x + y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Differentiate wrt $x$ a second time (applying product rule):

$8 + \left(y\right) \left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right) + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\therefore 8 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0$

and from the earlier equation:

$y \frac{\mathrm{dy}}{\mathrm{dx}} = - 8 x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 x}{y}$

Substituting gives us:

$8 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(\frac{- 8 x}{y}\right)}^{2} = 0$
$\therefore 8 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + \frac{64 {x}^{2}}{y} ^ 2 = 0$
$\therefore y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 8 - \frac{64 {x}^{2}}{y} ^ 2$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{8}{y} - \frac{64 {x}^{2}}{y} ^ 3$