Answer
$h'\left( x \right) = (2x^2 + 1)e^{x^2}$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {e^{{x^2} + \ln x}} \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{{x^2} + \ln x}}} \right] \cr
& {\text{Use the formula }}\frac{d}{{dx}}\left[ {{e^x}} \right]{\text{ and the chain rule}}{\text{, then}} \cr
& \frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}},{\text{ let }}u = {x^2} + \ln x \cr
& h'\left( x \right) = {e^{{x^2} + \ln x}}\frac{d}{{dx}}\left[ {{x^2} + \ln x} \right] \cr
& {\text{Computing derivatives}} \cr
& h'\left( x \right) = {e^{{x^2} + \ln x}}\left( {2x + \frac{1}{x}} \right) \cr
& {\text{Simplify}} \cr
& h'\left( x \right) = {e^{{x^2} + \ln x}}\left( {\frac{{2{x^2} + 1}}{x}} \right) \cr
& h'\left( x \right) = \left( {\frac{{2{x^2} + 1}}{x}} \right){e^{{x^2} + \ln x}} \cr
& h'\left( x \right) = \left( {\frac{{2{x^2} + 1}}{x}} \right){e^{{x^2}}\cdot e^{\ln x}} \cr
& h'\left( x \right) = \left( {\frac{{2{x^2} + 1}}{x}} \right){e^{{x^2}}\cdot x} \cr
& h'\left( x \right) = (2x^2 + 1)e^{x^2} \cr} $$
