Answer
$$k'\left( x \right) = \frac{{3{x^2} + 14}}{{x\left( {{x^2} + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& k\left( x \right) = 5\ln \left( {2{x^2} + 4} \right) - 7x \cr
& {\text{Calculate the derivative of the function}} \cr
& k'\left( x \right) = \frac{d}{{dx}}\left( {5\ln \left( {2{x^2} + 4} \right)} \right) - \frac{d}{{dx}}\left( {7x} \right) \cr
& {\text{Drop out the constants}} \cr
& k'\left( x \right) = 5\frac{d}{{dx}}\left( {\ln \left( {2{x^2} + 4} \right)} \right) - 7\frac{d}{{dx}}\left( x \right) \cr
& {\text{Compute derivatives}} \cr
& k'\left( x \right) = 5\left( {\frac{{\left( {2{x^2} + 4} \right)'}}{{2{x^2} + 4}}} \right) - 7\left( {\frac{1}{x}} \right) \cr
& k'\left( x \right) = 5\left( {\frac{{4x}}{{2{x^2} + 4}}} \right) - \frac{7}{x} \cr
& {\text{simplifying}} \cr
& k'\left( x \right) = 5\left( {\frac{{2x}}{{{x^2} + 2}}} \right) - \frac{7}{x} \cr
& k'\left( x \right) = \frac{{10x}}{{{x^2} + 2}} - \frac{7}{x} \cr
& k'\left( x \right) = \frac{{10{x^2} - 7{x^2} + 14}}{{x\left( {{x^2} + 2} \right)}} \cr
& k'\left( x \right) = \frac{{3{x^2} + 14}}{{x\left( {{x^2} + 2} \right)}} \cr} $$
