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

Feb 28, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 {y}^{2} - {x}^{2}}{4 {y}^{3}}$

Explanation:

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

$\Rightarrow 0 = 4 x - 8 y . \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 x}{- 8 y} = \frac{x}{2 y}$

The second derivative is obtained by differentiating $\frac{\mathrm{dy}}{\mathrm{dx}}$

differentiate $\frac{\mathrm{dy}}{\mathrm{dx}} \text{ using the " color(blue)"quotient rule}$

$\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y .1 - x .2 \frac{\mathrm{dy}}{\mathrm{dx}}}{4 {y}^{2}}$

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

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

$\textcolor{w h i t e}{\Rightarrow \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}} = \frac{2 {y}^{2} - {x}^{2}}{4 {y}^{3}}$