Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 8

$$\,\,\,\,\,\,\frac{{dw}}{{dt}} = \frac{3}{{3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}}} - \frac{4}{{3{t^{1/3}}\left( {3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}} \right)}} - \frac{{24{t^{ - 7}}}}{{3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}}}$$

$$\eqalign{ & w = \ln \left( {3{x^2} - 2y + 4{z^3}} \right);\,\,\,\,\,x = {t^{1/2}},\,\,\,y = {t^{2/3}},\,\,\,z = {t^{ - 2}} \cr & {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial x}}{\text{,}}\,\,\frac{{\partial w}}{{\partial y}}{\text{ and }}\frac{{\partial w}}{{\partial z}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\ln \left( {3{x^2} - 2y + 4{z^3}} \right)} \right) \cr & \,\,\,\,{\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{1}{{3{x^2} - 2y + 4{z^3}}}\frac{\partial }{{\partial x}}\left( {3{x^2} - 2y + 4{z^3}} \right) \cr & \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{{6x}}{{3{x^2} - 2y + 4{z^3}}} \cr & \cr & \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\ln \left( {3{x^2} - 2y + 4{z^3}} \right)} \right) \cr & \,\,\,\,{\text{treat }}x{\text{ and }}z{\text{ as constants}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{1}{{3{x^2} - 2y + 4{z^3}}}\frac{\partial }{{\partial y}}\left( {3{x^2} - 2y + 4{z^3}} \right) \cr & \,\,\,\,\frac{{\partial w}}{{\partial y}} = - \frac{2}{{3{x^2} - 2y + 4{z^3}}} \cr & \cr & \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {\ln \left( {3{x^2} - 2y + 4{z^3}} \right)} \right) \cr & \,\,\,\,{\text{treat }}x{\text{ and }}y{\text{ as constants}} \cr & \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{1}{{3{x^2} - 2y + 4{z^3}}}\frac{\partial }{{\partial z}}\left( {3{x^2} - 2y + 4{z^3}} \right) \cr & \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{{12{z^2}}}{{3{x^2} - 2y + 4{z^3}}} \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{,}}\,\,{\text{ }}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dz}}{{dt}} \cr & \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {{t^{1/2}}} \right) = \frac{1}{2}{t^{ - 1/2}} \cr & \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{t^{2/3}}} \right) = \frac{2}{3}{t^{ - 1/3}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{d}{{dt}}\left( {{t^{ - 2}}} \right) = - 2{t^{ - 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{{dw}}{{dt}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial w}}{{\partial y}}\frac{{dz}}{{dt}} \cr & \,\,\,\,\,{\text{Substitute the derivatives}} \cr & \,\,\,\,\frac{{dw}}{{dt}} = \left( {\frac{{6x}}{{3{x^2} - 2y + 4{z^3}}}} \right)\left( {\frac{1}{2}{t^{ - 1/2}}} \right) + \left( { - \frac{2}{{3{x^2} - 2y + 4{z^3}}}} \right)\left( {\frac{2}{3}{t^{ - 1/3}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {\frac{{12{z^2}}}{{3{x^2} - 2y + 4{z^3}}}} \right)\left( { - 2{t^{ - 3}}} \right) \cr & \,\,\,\,{\text{Where }}\,x = {t^{1/2}},\,\,\,y = {t^{2/3}},\,\,\,z = {t^{ - 2}} \cr & \,\,\,\,\frac{{dw}}{{dt}} = \left( {\frac{{6{t^{1/2}}}}{{3{{\left( {{t^{1/2}}} \right)}^2} - 2{t^{2/3}} + 4{{\left( {{t^{ - 2}}} \right)}^3}}}} \right)\left( {\frac{1}{2}{t^{ - 1/2}}} \right) + \left( { - \frac{2}{{3{{\left( {{t^{1/2}}} \right)}^2} - 2{t^{2/3}} + 4{{\left( {{t^{ - 2}}} \right)}^3}}}} \right)\left( {\frac{2}{3}{t^{ - 1/3}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {\frac{{12{{\left( {{t^{ - 2}}} \right)}^2}}}{{3{{\left( {{t^{1/2}}} \right)}^2} - 2{t^{2/3}} + 4{{\left( {{t^{ - 2}}} \right)}^3}}}} \right)\left( { - 2{t^{ - 3}}} \right) \cr & \,\,\,\,\,{\text{simplifying}} \cr & \,\,\,\,\,\,\frac{{dw}}{{dt}} = \frac{3}{{3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}}} - \frac{4}{{3{t^{1/3}}\left( {3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}} \right)}} - \frac{{24{t^{ - 7}}}}{{3{t^2} - 2{t^{2/3}} + 4{t^{ - 6}}}} \cr} $$