Answer
$$h'\left( x \right) = \left( {2x - 1} \right){e^{{x^2} - x + 1}}$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {e^{{x^2} - x + 1}} \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{{x^2} - x + 1}}} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}} \cr
& h'\left( x \right) = {e^{{x^2} - x + 1}}\frac{d}{{dx}}\left[ {{x^2} - x + 1} \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& h'\left( x \right) = {e^{{x^2} - x + 1}}\left( {2x - 1} \right) \cr
& {\text{Simplifying}} \cr
& h'\left( x \right) = \left( {2x - 1} \right){e^{{x^2} - x + 1}} \cr} $$
