Answer
$$\frac{{dy}}{{dt}} = \frac{{7{t^6}}}{{\sqrt {2{t^7} - 5} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {2{t^7} - 5} \cr
& {\text{write the radical as }}{\left( {2{t^7} - 5} \right)^{1/2}} \cr
& y = {\left( {2{t^7} - 5} \right)^{1/2}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {2{t^7} - 5} \right)}^{1/2}}} \right] \cr
& {\text{use the power rule with the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{2}{\left( {2{t^7} - 5} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {2{t^7} - 5} \right] \cr
& {\text{find derivative using }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{x - 1}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{2}{\left( {2{t^7} - 5} \right)^{ - 1/2}}\left( {14{t^6}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dt}} = 7{t^6}{\left( {2{t^7} - 5} \right)^{ - 1/2}} \cr
& \frac{{dy}}{{dt}} = \frac{{7{t^6}}}{{{{\left( {2{t^7} - 5} \right)}^{ - 1/2}}}} \cr
& \frac{{dy}}{{dt}} = \frac{{7{t^6}}}{{\sqrt {2{t^7} - 5} }} \cr} $$
