Answer
$$f'\left( x \right) = \left[ {29\left( {{{1.7}^x}} \right)} \right]\left( {25.6x + 3.7 + \left( {12.8{x^2} + 3.7x + 1.2} \right)\left( {\ln 1.7} \right)} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \left( {12.8{x^2} + 3.7x + 1.2} \right)\left[ {29\left( {{{1.7}^x}} \right)} \right] \cr
& {\text{Calculate the derivative of the function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {\left( {12.8{x^2} + 3.7x + 1.2} \right)\left[ {29\left( {{{1.7}^x}} \right)} \right]} \right) \cr
& {\text{use the product rule }} \cr
& f'\left( x \right) = \left[ {29\left( {{{1.7}^x}} \right)} \right]\frac{d}{{dx}}\left( {12.8{x^2} + 3.7x + 1.2} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {12.8{x^2} + 3.7x + 1.2} \right)\frac{d}{{dx}}\left[ {29\left( {{{1.7}^x}} \right)} \right] \cr
& {\text{use the power rule }}\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}{\text{ and }}\frac{d}{{dx}}\left[ {{a^x}} \right] = {a^x}\ln a \cr
& f'\left( x \right) = \left[ {29\left( {{{1.7}^x}} \right)} \right]\left( {25.6x + 3.7} \right) + \left( {12.8{x^2} + 3.7x + 1.2} \right)\left( {29\left( {{{1.7}^x}} \right)\left( {\ln 1.7} \right)} \right) \cr
& {\text{Factoring }}29\left( {{{1.7}^x}} \right) \cr
& f'\left( x \right) = \left[ {29\left( {{{1.7}^x}} \right)} \right]\left( {25.6x + 3.7 + \left( {12.8{x^2} + 3.7x + 1.2} \right)\left( {\ln 1.7} \right)} \right) \cr} $$
