## Calculus: Early Transcendentals (2nd Edition)

$\frac{{{d^3}}}{{d{x^3}}} = 29.568$
$\begin{gathered} {\left. {\frac{{{d^3}}}{{d{x^3}}}\,\left( {{x^{4.2}}} \right)} \right|_{x = 1}} \hfill \\ \hfill \\ differentiate\,\,use\,\,the\,\,power\,\,rule \hfill \\ \hfill \\ \frac{d}{{dx}} = 4.2{x^{4.2 - 1}} = 4.2{x^{3.2}} \hfill \\ \hfill \\ differentiate\,\,use\,\,the\,\,power\,\,rule{\text{ again}} \hfill \\ \hfill \\ \frac{{{d^2}}}{{d{x^2}}} = \,\left( {4.2} \right)\,\left( {3.2} \right){x^{3.2 - 1}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{{d^2}}}{{d{x^2}}} = 13.44{x^{2.2}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{{d^3}}}{{d{x^3}}} = \,\left( {2.2} \right)\,\left( {13.44} \right){x^{2.2 - 1}} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ \frac{{{d^3}}}{{d{x^3}}} = 29.568{x^{1.1}} \hfill \\ \hfill \\ Evaluate\,\,\,{\text{ at}}\;\;{\text{ }}x = 1 \hfill \\ \hfill \\ \frac{{{d^3}}}{{d{x^3}}} = 29.568{\left( 1 \right)^{1.1}} \hfill \\ \hfill \\ \frac{{{d^3}}}{{d{x^3}}} = 29.568 \hfill \\ \end{gathered}$