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

Feb 19, 2015

You can calculate a derivative of a function, but $x 2 {y}^{2} - 1$ is not a function, it's an equality...

Maybe you're looking for derivative $f \left(x , y\right) = {x}^{2} {y}^{2} - 1$ ? But for which variable ? $x$ or $y$ ?

Feb 19, 2015

First we need to find the first derivative of the function.
We will do this using implicit differentiation. With this
particular function we will use the product rule.

$2 x {y}^{2} + {x}^{2} \left(2 y\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Now subtract $2 x {y}^{2}$ from both sides

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

Divide both sides by ${x}^{2} \left(2 y\right)$

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

Which simplifies to

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

Now for the second derivative we will use the quotient rule

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\left(x \left(- 1\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left(1 \left(- y\right)\right)}{x} ^ 2$

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

plug $\frac{\mathrm{dy}}{\mathrm{dx}}$ into the right hand side

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

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