# How do you find the second derivative of 2x^2-y^2=1?

May 5, 2017

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

#### Explanation:

We have
$2 {x}^{2} - {y}^{2} = 1$.........equation 1
Differentiating with respect to $x$ we get
$4 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \frac{x}{y}$
Again differentiating with respect to $x$ we get
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 \frac{\left(y - x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{y} ^ 2$
Now putting the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ we get
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 \frac{\left({y}^{2} - 2 {x}^{2}\right)}{y} ^ 3$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2}{y} ^ 3$ (using equation 1)