#### Answer

\[{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right) \cdot {f^,}\,\left( {g\,\left( {{x^m}} \right)} \right) \cdot {g^,}\,\left( {{x^m}} \right) \cdot {mx^{m - 1}}\]

#### Work Step by Step

\[\begin{gathered}
Use\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
f\,{\left( {g\,\left( t \right)} \right)^,} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right)\, \cdot \,{\left( {f\,\left( {g\,\left( {{x^m}} \right)} \right)} \right)^,} \hfill \\
\hfill \\
Multiply \hfill \\
\hfill \\
{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right) \cdot {f^,}\,\left( {g\,\left( {{x^m}} \right)} \right) \cdot \,{\left( {g\,\left( {{x^m}} \right)} \right)^,} \hfill \\
\hfill \\
{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right) \cdot {f^,}\,\left( {g\,\left( {{x^m}} \right)} \right) \cdot {g^,}\,\left( {{x^m}} \right) \cdot \,{\left( {{x^m}} \right)^,} \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right) \cdot {f^,}\,\left( {g\,\left( {{x^m}} \right)} \right) \cdot {g^,}\,\left( {{x^m}} \right) \cdot m{x^{m - 1}} \hfill \\
\hfill \\
{y^,} = n{f^{n - 1}}\,\left( {g\left( {{x^m}} \right)} \right) \cdot {f^,}\,\left( {g\,\left( {{x^m}} \right)} \right) \cdot {g^,}\,\left( {{x^m}} \right) \cdot {mx^{m - 1}} \hfill \\
\hfill \\
\end{gathered} \]