Answer
$$\frac{{dy}}{{dt}} = - \frac{{48{t^3}}}{{\sqrt {8{t^4} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = - 3\sqrt {8{t^4} - 1} \cr
& {\text{write the radical as }}{\left( {8{t^4} - 1} \right)^{1/2}} \cr
& y = - 3{\left( {8{t^4} - 1} \right)^{1/2}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ { - 3{{\left( {8{t^4} - 1} \right)}^{1/2}}} \right] \cr
& {\text{use the power rule with the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr
& \frac{{dy}}{{dt}} = - 3\left( {\frac{1}{2}} \right){\left( {8{t^4} - 1} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {8{t^4} - 1} \right] \cr
& {\text{find derivative using }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{x - 1}} \cr
& \frac{{dy}}{{dt}} = - 3\left( {\frac{1}{2}} \right){\left( {8{t^4} - 1} \right)^{ - 1/2}}\left( {32{t^3}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dt}} = - 3\left( {16{t^3}} \right){\left( {8{t^4} - 1} \right)^{ - 1/2}} \cr
& \frac{{dy}}{{dt}} = - \frac{{48{t^3}}}{{{{\left( {8{t^4} - 1} \right)}^{1/2}}}} \cr
& \frac{{dy}}{{dt}} = - \frac{{48{t^3}}}{{\sqrt {8{t^4} - 1} }} \cr} $$
