Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 956: 2

$$\,\,\,\,\frac{{dz}}{{dt}} = \frac{2}{{2t + {t^{2/3}}}} + \frac{2}{{6{t^{4/3}} + 3t}}$$

$$\eqalign{ & z = \ln \left( {2{x^2} + y} \right);\,\,\,\,\,\,\,\,\,x = \sqrt t ,\,\,\,\,\,\,\,y = {t^{2/3}} \cr & \cr & {\text{Calculate the partial derivatives }}\frac{{\partial z}}{{\partial x}}{\text{ and }}\frac{{\partial y}}{{\partial x}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\ln \left( {2{x^2} + y} \right)} \right) \cr & \,\,\,\,{\text{treat }}y{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{1}{{2{x^2} + y}}\frac{\partial }{{\partial x}}\left( {2{x^2} + y} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{1}{{2{x^2} + y}}\left( {4x} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{{4x}}{{2{x^2} + y}} \cr & and \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\ln \left( {2{x^2} + y} \right)} \right) \cr & \,\,\,\,{\text{treat }}x{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{1}{{2{x^2} + y}}\frac{\partial }{{\partial y}}\left( {2{x^2} + y} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{1}{{2{x^2} + y}}\left( 1 \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{1}{{2{x^2} + y}} \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\sqrt t } \right) = \frac{1}{{2\sqrt t }} \cr & \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^{2/3}}} \right) = \frac{2}{3}{t^{ - 1/3}} \cr & \cr & {\text{Use the Theorem 13}}{\text{.5}}{\text{.1}}\,\,\left( {{\text{chain rule for derivatives}}} \right){\text{ see page 950}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {\frac{{4x}}{{2{x^2} + y}}} \right)\left( {\frac{1}{{2\sqrt t }}} \right) + \left( {\frac{1}{{2{x^2} + y}}} \right)\left( {\frac{2}{3}{t^{ - 1/3}}} \right) \cr & \,\,\,\,{\text{Where }}x = \sqrt t {\text{ and}}\,\,\,\,\,y = {t^{2/3}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {\frac{{4\sqrt t }}{{2{{\left( {\sqrt t } \right)}^2} + {t^{2/3}}}}} \right)\left( {\frac{1}{{2\sqrt t }}} \right) + \left( {\frac{1}{{2{{\left( {\sqrt t } \right)}^2} + {t^{2/3}}}}} \right)\left( {\frac{2}{3}{t^{ - 1/3}}} \right) \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{2}{{2t + {t^{2/3}}}} + \frac{2}{{6{t^{4/3}} + 3t}} \cr} $$