Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 956: 4

$$\,\,\,\,\frac{{dz}}{{dt}} = \frac{{1 - 2{t^4}}}{{2t\sqrt {1 + \ln t - 2{t^4}\ln t} }} - \frac{{4{t^3}\ln t}}{{2\sqrt {1 + \ln t - 2{t^4}\ln t} }}$$

$$\eqalign{ & z = \sqrt {1 + x - 2x{y^4}} ;\,\,\,\,\,\,\,\,\,x = \ln t,\,\,\,\,\,\,\,y = t \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( {\sqrt {1 + x - 2x{y^4}} } \right) \cr & \,\,\,\,{\text{treat }}y{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{1}{2}{\left( {1 + x - 2x{y^4}} \right)^{ - 1/2}}\left( {1 - 2{y^4}} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial x}} = \frac{{1 - 2{y^4}}}{{2\sqrt {1 + x - 2x{y^4}} }} \cr & and \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\sqrt {1 + x - 2x{y^4}} } \right) \cr & \,\,\,\,{\text{treat }}x{\text{ as a constant}} \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{1}{2}{\left( {1 + x - 2x{y^4}} \right)^{ - 1/2}}\left( { - 8x{y^3}} \right) \cr & \,\,\,\,\frac{{\partial z}}{{\partial y}} = \frac{{ - 4x{y^3}}}{{\sqrt {1 + x - 2x{y^4}} }} \cr & \cr & {\text{Calculate the ordinary derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \,\,\,\,\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\ln t} \right) = \frac{1}{t} \cr & \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1 \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{{1 - 2{y^4}}}{{2\sqrt {1 + x - 2x{y^4}} }}} \right)\left( {\frac{1}{t}} \right) + \left( {\frac{{ - 4x{y^3}}}{{\sqrt {1 + x - 2x{y^4}} }}} \right)\left( 1 \right) \cr & \,\,\,\,{\text{Where }}x = \ln t{\text{ and }}\,\,\,\,\,y = t \cr & \,\,\,\,\frac{{dz}}{{dt}} = \left( {\frac{{1 - 2{{\left( t \right)}^4}}}{{2\sqrt {1 + \ln t - 2\left( {\ln t} \right){{\left( t \right)}^4}} }}} \right)\left( {\frac{1}{t}} \right) + \left( {\frac{{ - 4\left( {\ln t} \right){{\left( t \right)}^3}}}{{\sqrt {1 + \ln t - 2\left( {\ln t} \right){{\left( t \right)}^4}} }}} \right)\left( 1 \right) \cr & \,\,\,\,\,{\text{simplifying}} \cr & \,\,\,\,\frac{{dz}}{{dt}} = \frac{{1 - 2{t^4}}}{{2t\sqrt {1 + \ln t - 2{t^4}\ln t} }} - \frac{{4{t^3}\ln t}}{{2\sqrt {1 + \ln t - 2{t^4}\ln t} }} \cr} $$