# How do you find the second derivative of 4x^2 +9y^2 = 36?

Feb 28, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{16}{9 {y}^{3}}$

#### Explanation:

differentiate all terms on both sides $\textcolor{b l u e}{\text{implicitly with respect to x}}$

$\Rightarrow 8 x + 18 y . \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 8 x}{18 y} = - \frac{4 x}{9 y}$

$\text{to obtain "(d^2y)/(dx^2)" differentiate " dy/dx" using the "color(blue)" quotient rule}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{9 y . \left(- 4\right) - \left(- 4 x\right) . \left(9 \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{81 {y}^{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{- 36 y + 36 x . \frac{\mathrm{dy}}{\mathrm{dx}}}{81 {y}^{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{- 36 y + 36 x . \left(- \frac{4 x}{9 y}\right)}{81 {y}^{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{- 36 y - \frac{16 {x}^{2}}{y}}{81 {y}^{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{- 36 {y}^{2} - 16 {x}^{2}}{81 {y}^{3}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{- 4 \left(9 {y}^{2} + 4 {x}^{2}\right)}{81 {y}^{3}}$

$\text{now "9y^2+4x^2=36larr" initial statement}$

$\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{144}{81 {y}^{3}}$

$\textcolor{w h i t e}{\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = - \frac{16}{9 {y}^{3}}$