Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises: 47

Answer

\[{g^,}\,\left( z \right) = \,\,\frac{{3{z^2} + 1}}{{\,\left( {\ln 2} \right)\,\left( {{z^3} + z + 1} \right)}}\]

Work Step by Step

\[\begin{gathered} g\,\left( z \right) = {\log _2}\,\left( {{z^3} + z + 1} \right) \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\ {g^,}\,\left( z \right) = \,\,{\left[ {{{\log }_2}\,\left( {{z^3} + z + 1} \right)} \right]^,} \hfill \\ Use\,\,the\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\left[ {{{\log }_a}g\,\left( x \right)} \right] = \frac{1}{{\ln a}}\,\left( {\frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}}} \right)\, \hfill \\ Then \hfill \\ {g^,}\,\left( z \right) = \,\,\,\left( {\frac{1}{{\ln 2}}} \right)\,\left( {\frac{{\,{{\left( {{z^3} + z + 1} \right)}^,}}}{{{z^3} + z + 1}}} \right) \hfill \\ {g^,}\,\left( z \right) = \,\,\left( {\frac{1}{{\ln 2}}} \right)\,\left( {\frac{{3{z^2} + 1}}{{{z^3} + z + 1}}} \right) \hfill \\ {g^,}\,\left( z \right) = \,\,\frac{{3{z^2} + 1}}{{\,\left( {\ln 2} \right)\,\left( {{z^3} + z + 1} \right)}} \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.