# Chapter 3: Derivatives - Practice Exercises - Page 178: 62

$${\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {8,8} \right)}} = \frac{1}{6}$$

#### Work Step by Step

\eqalign{ & {x^{1/3}} + {y^{1/3}} = 4 \cr & \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^{1/3}}} \right) + \frac{d}{{dx}}\left( {{y^{1/3}}} \right) = \frac{d}{{dx}}\left( 4 \right) \cr & {\text{find derivatives}} \cr & \frac{1}{3}{x^{ - 2/3}} + \frac{1}{3}{y^{ - 2/3}}\frac{{dy}}{{dx}} = 0 \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \frac{1}{3}{y^{ - 2/3}}\frac{{dy}}{{dx}} = - \frac{1}{3}{x^{ - 2/3}} \cr & \frac{{dy}}{{dx}} = - \frac{{{x^{ - 2/3}}}}{{{y^{ - 2/3}}}} \cr & \cr & {\text{find the second derivative}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{d}{{dx}}\left( {\frac{{{x^{ - 2/3}}}}{{{y^{ - 2/3}}}}} \right) \cr & {\text{use the quotient rule}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^{ - 2/3}}\frac{d}{{dx}}\left[ {{x^{ - 2/3}}} \right] - {x^{ - 2/3}}\frac{d}{{dx}}\left[ {{y^{ - 2/3}}} \right]}}{{{{\left( {{y^{ - 2/3}}} \right)}^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^{ - 2/3}}\left( { - \frac{2}{3}{x^{ - 5/3}}} \right) - {x^{ - 2/3}}\left( { - \frac{2}{3}{y^{ - 5/3}}} \right)\frac{{dy}}{{dx}}}}{{{y^{ - 4/3}}}} \cr & \cr & {\text{substitute }}\frac{{dy}}{{dx}} = - \frac{{{x^{ - 2/3}}}}{{{y^{ - 2/3}}}}{\text{ and simplify}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{y^{ - 2/3}}\left( { - \frac{2}{3}{x^{ - 5/3}}} \right) - {x^{ - 2/3}}\left( { - \frac{2}{3}{y^{ - 5/3}}} \right)\left( { - \frac{{{x^{ - 2/3}}}}{{{y^{ - 2/3}}}}} \right)}}{{{y^{ - 4/3}}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{{ - \frac{2}{3}{x^{ - 5/3}}{y^{ - 2/3}} - \frac{2}{3}{x^{ - 4/3}}{y^{ - 1}}}}{{{y^{ - 4/3}}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{2}{3}{x^{ - 5/3}}{y^{2/3}} + \frac{2}{3}{x^{ - 4/3}}{y^{1/3}} \cr & \cr & {\text{Evaluate at the point }}\left( {8,8} \right) \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {8,8} \right)}} = \frac{2}{3}{\left( 8 \right)^{ - 5/3}}{\left( 8 \right)^{2/3}} + \frac{2}{3}{\left( 8 \right)^{ - 4/3}}{\left( 8 \right)^{1/3}} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\left( {8,8} \right)}} = \frac{1}{6} \cr}

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