## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = - \frac{x}{y}{\text{ and }}\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^2} + {x^2}}}{{{y^3}}}=\frac{-1}{y^3}$$
\eqalign{ & {x^2} + {y^2} = 1 \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr & \cr & {\text{find derivatives}} \cr & 2x + 2y\frac{{dy}}{{dx}} = 0 \cr & \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & 2y\frac{{dy}}{{dx}} = - 2x \cr & \frac{{dy}}{{dx}} = - \frac{x}{y} \cr & \cr & {\text{find the second derivative}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( { - \frac{x}{y}} \right) \cr & {\text{use the quotient rule}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{y\left( 1 \right) - x\frac{d}{{dx}}\left( y \right)}}{{{y^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{y - x\frac{{dy}}{{dx}}}}{{{y^2}}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = - \frac{x}{y}{\text{ and simplify}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{y - x\left( { - \frac{x}{y}} \right)}}{{{y^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{y + \frac{{{x^2}}}{y}}}{{{y^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^2} + {x^2}}}{{{y^3}}}=-\frac{1}{y^3} \cr}