#### Answer

\[{y^,} = \frac{{{e^{{x^2}}}}}{x} + 2x{e^{{x^2}}}\ln x\]

#### Work Step by Step

\[\begin{gathered}
y = {e^{{x^2}}}\ln x \hfill \\
Differentiate \hfill \\
{y^,} = \,\,\left[ {{e^{{x^2}}}\ln x} \right] \hfill \\
Use\,\,the\,\,product\,\,rule\, \hfill \\
{y^,} = {e^{{x^2}}}\,{\left( {\ln x} \right)^,} + \ln x\,\left( {{e^{{x^2}}}} \right) \hfill \\
Use\,\,the\,\,\,formula \hfill \\
\frac{d}{{dx}}\,\,\,\,\left[ {\ln x} \right] = \frac{1}{x} \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{y^,} = {e^{{x^2}}}\,\left( {\frac{1}{x}} \right) + \ln x\,\left( {2x{e^{{x^2}}}} \right) \hfill \\
Multipliying \hfill \\
{y^,} = \frac{{{e^{{x^2}}}}}{x} + 2x{e^{{x^2}}}\ln x \hfill \\
\end{gathered} \]