Answer
$$\,\frac{{dw}}{{dt}} = \frac{{1 - 512{t^5}}}{{2t\sqrt {1 + \ln t - 512{t^5}\ln t} }} - \frac{{256{t^4}\ln t}}{{\sqrt {1 + \ln t - 512{t^5}\ln t} }} - \frac{{1024{t^4}\ln t}}{{\sqrt {1 + \ln t - 512{t^5}\ln t} }}$$
Work Step by Step
$$\eqalign{
& w = \sqrt {1 + x - 2y{z^4}x} ;\,\,\,\,\,x = \ln t,\,\,\,y = t,\,\,\,z = 4t \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( {\sqrt {1 + x - 2y{z^4}x} } \right) \cr
& \,\,\,\,{\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{1}{2}{\left( {1 + x - 2y{z^4}x} \right)^{ - 1/2}}\left( {1 - 2y{z^4}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial x}} = \frac{{1 - 2y{z^4}}}{{2\sqrt {1 + x - 2y{z^4}x} }} \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\sqrt {1 + x - 2y{z^4}x} } \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}z{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = \frac{1}{2}{\left( {1 + x - 2y{z^4}x} \right)^{ - 1/2}}\left( { - 2x{z^4}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial y}} = - \frac{{x{z^4}}}{{\sqrt {1 + x - 2y{z^4}x} }} \cr
& \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {\sqrt {1 + x - 2y{z^4}x} } \right) \cr
& \,\,\,\,{\text{treat }}x{\text{ and }}y{\text{ as constants}} \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = \frac{1}{2}{\left( {1 + x - 2y{z^4}x} \right)^{ - 1/2}}\left( { - 8xy{z^3}} \right) \cr
& \,\,\,\,\frac{{\partial w}}{{\partial z}} = - \frac{{4xy{z^3}}}{{\sqrt {1 + x - 2y{z^4}x} }} \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( {\ln t} \right) = \frac{1}{t} \cr
& \,\,\,\,\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( t \right) = 1 \cr
& \,\,\,\,\frac{{dz}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4 \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{{1 - 2y{z^4}}}{{2\sqrt {1 + x - 2y{z^4}x} }}} \right)\left( {\frac{1}{t}} \right) + \left( { - \frac{{x{z^4}}}{{\sqrt {1 + x - 2y{z^4}x} }}} \right) + \left( { - \frac{{4xy{z^3}}}{{\sqrt {1 + x - 2y{z^4}x} }}} \right)\left( 4 \right) \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \frac{{1 - 2y{z^4}}}{{2t\sqrt {1 + x - 2y{z^4}x} }} - \frac{{x{z^4}}}{{\sqrt {1 + x - 2y{z^4}x} }} - \frac{{16xy{z^3}}}{{\sqrt {1 + x - 2y{z^4}x} }} \cr
& \,\,\,\,{\text{Where }}\,x = \ln t,\,\,\,y = t,\,\,\,z = 4t \cr
& \,\,\,\,\frac{{dw}}{{dt}} = \frac{{1 - 2\left( t \right){{\left( {4t} \right)}^4}}}{{2t\sqrt {1 + \ln t - 2t{{\left( {4t} \right)}^4}\ln t} }} - \frac{{{{\left( {4t} \right)}^4}\ln t}}{{\sqrt {1 + \ln t - 2t{{\left( {4t} \right)}^4}\ln t} }} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{{16t{{\left( {4t} \right)}^3}\ln t}}{{\sqrt {1 + \ln t - 2t{{\left( {4t} \right)}^4}\ln t} }} \cr
& \,\frac{{dw}}{{dt}} = \frac{{1 - 512{t^5}}}{{2t\sqrt {1 + \ln t - 512{t^5}\ln t} }} - \frac{{256{t^4}\ln t}}{{\sqrt {1 + \ln t - 512{t^5}\ln t} }} - \frac{{1024{t^4}\ln t}}{{\sqrt {1 + \ln t - 512{t^5}\ln t} }} \cr} $$
