Answer
\[{f^,}\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}}\,\,\left( {\frac{{4x - {x^4}}}{{\,{{\left( {{x^3} + 2} \right)}^2}}}} \right)\]
Work Step by Step
\[\begin{gathered}
f\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}} \hfill \\
Differentiate\,\,using\,\,the\,formula \hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}}\,\,{\left[ {\frac{{{x^2}}}{{{x^3} + 2}}} \right]^,} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{f^,}\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}}\,\,\left( {\frac{{\,\left( {{x^3} + 2} \right)\,{{\left( {{x^2}} \right)}^,} - {x^2}\,{{\left( {{x^3} + 2} \right)}^,}}}{{\,{{\left( {{x^3} + 2} \right)}^2}}}} \right) \hfill \\
Then \hfill \\
{f^,}\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}}\,\left( {\frac{{\left( {{x^3} + 2} \right)\,\left( {2x} \right) - {x^2}\,\left( {3{x^2}} \right)}}{{\,{{\left( {{x^3} + 2} \right)}^2}}}} \right) \hfill \\
Multiplying \hfill \\
\hfill \\
{f^,}\,\left( x \right) = {e^{\frac{{{x^2}}}{{{x^3} + 2}}}}\,\,\left( {\frac{{4x - {x^4}}}{{\,{{\left( {{x^3} + 2} \right)}^2}}}} \right) \hfill \\
\end{gathered} \]
