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

Nov 6, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left[{x}^{2} + 4 {y}^{2} = 5\right] = \frac{- 5}{16 {y}^{3}}$

#### Explanation:

First find $\frac{\mathrm{dy}}{\mathrm{dx}}$ by implicitly differentiating ${x}^{2} + 4 {y}^{2} = 5$:
${x}^{2} + 4 {y}^{2} = 5$
$2 x + 8 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(8 y\right) = - 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x}{8 y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x}{4 y}$

Now implicitly differentiate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

Use the quotient rule:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(4 y\right) \left(- 1\right) - \left(- x\right) \left(4 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)}{{\left(4 y\right)}^{2}}$

Simplify:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 4 y + 4 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{16 {y}^{2}}$

Substitute $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x}{4 y}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- \cancel{4} y + \cancel{4} x \left(\frac{- x}{4 y}\right)}{4 \cancel{16} {y}^{2}}$

Simplify:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left[\frac{- y}{4 {y}^{2}}\right] - \left[\frac{{x}^{2}}{\left(4 y\right) \left(4 {y}^{2}\right)}\right]$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left[\frac{- 1}{4 y}\right] - \left[\frac{{x}^{2}}{16 {y}^{3}}\right]$

Make a common denominator to combine into one fraction:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 4 {y}^{2} - {x}^{2}}{16 {y}^{3}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- \left(4 {y}^{2} + {x}^{2}\right)}{16 {y}^{3}}$

Recall that the original equation states that $4 {y}^{2} + {x}^{2} = 5$, so:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 5}{16 {y}^{3}}$