Answer
$$h'\left( x \right) = 2\left( {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right)\left( {6.2x + \frac{{3.1}}{{{{\left( {3.1x - 2} \right)}^2}}}} \right)$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {\left[ {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right]^2} \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left( {{{\left[ {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right]}^2}} \right) \cr
& {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& h'\left( x \right) = 2\left[ {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right]\frac{d}{{dx}}\left( {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right) \cr
& {\text{Computing derivatives}} \cr
& h'\left( x \right) = 2\left( {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right)\left( {6.2x - 0 - \left( {\frac{{ - 3.1}}{{{{\left( {3.1x - 2} \right)}^2}}}} \right)} \right) \cr
& h'\left( x \right) = 2\left( {3.1{x^2} - 2 - \frac{1}{{3.1x - 2}}} \right)\left( {6.2x + \frac{{3.1}}{{{{\left( {3.1x - 2} \right)}^2}}}} \right) \cr} $$
