# How do you find (d^2y)/(dx^2) for 3x^2+y^2=2?

Feb 19, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{6}{y} ^ 3$

#### Explanation:

When we differentiate $y$, we get $\frac{\mathrm{dy}}{\mathrm{dx}}$ When we differentiate a non implicit function of $y$ then using the chain rule we can differentiate wrt $x$ and multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$. When this is done in situ it is known as implicit differentiation.

We have:

$3 {x}^{2} + {y}^{2} = 2$

Differentiate wrt $x$ and we get:

$\setminus \setminus \setminus 6 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 3 x + y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

And now (as we want the second derivative) we differentiate wrt x again, this time we must also apply the product rule:

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

We can rearrange the 1st derivative equation:

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 x}{y}$

And if we substitute this into the 2nd derivative equation we get:

$\therefore 3 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(- \frac{3 x}{y}\right)}^{2} = 0$
$\therefore 3 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + \frac{9 {x}^{2}}{y} ^ 2 = 0$
$\therefore y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{9 {x}^{2}}{y} ^ 2 - 3$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{9 {x}^{2}}{y} ^ 3 - \frac{3}{y}$
$\text{ } = - \frac{3 \left(3 {x}^{2} + {y}^{2}\right)}{y} ^ 3$
$\text{ } = - \frac{3 \left(2\right)}{y} ^ 3$
$\text{ } = - \frac{6}{y} ^ 3$

Feb 19, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \pm \frac{- 6}{2 - 3 {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

We can solve the equation to make $y \left(x\right)$ explicit:

$y \left(x\right) = \pm \sqrt{2 - 3 {x}^{2}}$

So:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \pm \frac{- 3 x}{\sqrt{2 - 3 {x}^{2}}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \pm \frac{- 3 \sqrt{2 - 3 {x}^{2}} - \left(\left(- 3 x\right) \frac{- 6 x}{2 \sqrt{2 - 3 {x}^{2}}}\right)}{2 - 3 {x}^{2}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \pm \frac{- 3 \left(2 - 3 {x}^{2}\right) - 9 {x}^{2}}{\left(2 - 3 {x}^{2}\right) \sqrt{2 - 3 {x}^{2}}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \pm \frac{- 6 + 9 {x}^{2} - 9 {x}^{2}}{\left(2 - 3 {x}^{2}\right) \sqrt{2 - 3 {x}^{2}}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \pm \frac{- 6}{\left(2 - 3 {x}^{2}\right) \sqrt{2 - 3 {x}^{2}}}$

We can also proceed implicitly by differentiating both sides of the equation with respect to $x$:

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

$6 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 x}{y}$

Differentiate again:

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

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

substituting $\frac{\mathrm{dy}}{\mathrm{dx}}$ from the equation above:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 9 {x}^{2}}{y} ^ 3 - \frac{3}{y}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{9 {x}^{2} + 3 {y}^{2}}{y} ^ 3$

and as ${y}^{2} = 2 - 3 {x}^{2}$:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{9 {x}^{2} + 6 - 9 {x}^{2}}{y} ^ 3$

simplifying:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{6}{y} ^ 3$

and substituting $y$ from the original equation we get the same result.