## University Calculus: Early Transcendentals (3rd Edition)

$\frac{d^2y}{dx}=\frac{1}{6}$
$y^{\frac{1}{3}}+x^{\frac{1}{3}}=4$ differentiate the above equation with respect to x: $\frac{1}{3}y^{\frac{-2}{3}}\frac{dy}{dx}=-\frac{1}{3}x^{\frac{-2}{3}}$ on taking double differentiation: $\frac{d^2y}{dx}=-\frac{\frac{-2}{9}x^{\frac{-5}{3}}{y^{\frac{-2}{3}}}+\frac{2}{9}y^{\frac{-5}{3}}{x^{\frac{-2}{3}}}\frac{dy}{dx}}{y^{\frac{-4}{3}}}$ at point$(8,8)$ $\frac{1}{3}8^{\frac{-2}{3}}\frac{dy}{dx}=-\frac{1}{3}8^{\frac{-2}{3}}$ $\frac{dy}{dx}=-1$ and $\frac{d^2y}{dx}=-\frac{\frac{-2}{9}8^{\frac{-5}{3}}{8^{\frac{-2}{3}}}+\frac{2}{9}8^{\frac{-5}{3}}{8^{\frac{-2}{3}}}{(-1)}}{8^{\frac{-4}{3}}}$ $\frac{d^2y}{dx}=\frac{1}{6}$