Answer
$$k'\left( x \right) = 90{x^2}{\left( {5{x^3} - 1} \right)^5}$$
Work Step by Step
$$\eqalign{
& k\left( x \right) = {\left( {5{x^3} - 1} \right)^6} \cr
& {\text{differentiate both sides}} \cr
& k'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {5{x^3} - 1} \right)}^6}} \right] \cr
& {\text{use the power rule with the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr
& k'\left( x \right) = 6{\left( {5{x^3} - 1} \right)^5}\frac{d}{{dx}}\left[ {5{x^3} - 1} \right] \cr
& {\text{find derivative using }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{x - 1}} \cr
& k'\left( x \right) = 6{\left( {5{x^3} - 1} \right)^5}\left( {15{x^2}} \right) \cr
& {\text{simplify}} \cr
& k'\left( x \right) = 90{x^2}{\left( {5{x^3} - 1} \right)^5} \cr} $$