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

Mar 9, 2018

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{y} - {x}^{2} / {y}^{3}$

#### Explanation:

We use implicit differentiation as follows:

differentiating ${x}^{2} + {y}^{2} = 1$, we get

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

and differentiating it further

$2 + 2 \left[\frac{\mathrm{dy}}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot \frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right\} = 0$

or $2 + 2 {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}^{2} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

or $2 + 2 {x}^{2} / {y}^{2} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

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

or $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2}{2 y} - 2 {x}^{2} / {y}^{2} \cdot \frac{1}{2 y}$

= $- \frac{1}{y} - {x}^{2} / {y}^{3}$