# How do you find y'' by implicit differentiation for 4x^2 + 3y^2 = 6?

Apr 16, 2015

$4 {x}^{2} + 3 {y}^{2} = 6$

$\frac{d}{\mathrm{dx}} \left(4 {x}^{2} + 3 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(6\right)$

$8 x + 6 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 x}{3 y}$

On to the second derivative:

$\frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left(\frac{- 4 x}{3 y}\right)$

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

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 12 y + 12 x \frac{\mathrm{dy}}{\mathrm{dx}}}{9 {y}^{2}}$

Replacing $\frac{\mathrm{dy}}{\mathrm{dx}}$ by the expression above gives us:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 12 y + 12 x \left(\frac{- 4 x}{3 y}\right)}{9 {y}^{2}} = \frac{- 12 y + \frac{- 16 {x}^{2}}{y}}{9 {y}^{2}}$

No multiply by $\frac{y}{y}$ to get

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 12 {y}^{2} - 16 {x}^{2}}{9 {y}^{3}} = \frac{- 4 \left(4 {x}^{2} + 3 {y}^{2}\right)}{9 {y}^{3}}$

Now, use the initial equation: $4 {x}^{2} + 3 {y}^{2} = 6$ to write the answer as:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = = \frac{- 24}{9 {y}^{3}} = \frac{- 8}{3 {y}^{3}}$