## Calculus: Early Transcendentals (2nd Edition)

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