Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 29

Answer

\[{g^,}\,\left( z \right) = \,\left( {6{e^{2z}} + \frac{3}{z}} \right)\,{\left( {{e^{2z}} + \ln z} \right)^2}\]

Work Step by Step

\[\begin{gathered} g\,\left( x \right) = \,{\left( {{e^{2z}} + \ln z} \right)^3} \hfill \\ Use\,\,the\,\,\,chain\,\,rule \hfill \\ {g^,}\,\left( z \right) = 3\,{\left( {{e^{2z}} + \ln z} \right)^{3 - 1}}\,{\left( {{e^{2z}} + \ln z} \right)^,} \hfill \\ Use\,\,the\,\,\,formula \hfill \\ \frac{d}{{dz}}\,\,{\left[ {{e^{g\,\left( z \right)}}} \right]^,} = {e^{g\,\left( z \right)}}{g^,}\,\left( z \right)\frac{d}{{dz}}\,\,\left[ {\ln z} \right] = \frac{1}{z} \hfill \\ Then \hfill \\ {g^,}\,\left( z \right) = 3\,{\left( {{e^{2z}} + \ln z} \right)^2}\,\left( {2{e^{2z}} + \frac{1}{z}} \right) \hfill \\ Multiplaying \hfill \\ {g^,}\,\left( z \right) = \,\left( {6{e^{2z}} + \frac{3}{z}} \right)\,{\left( {{e^{2z}} + \ln z} \right)^2} \hfill \\ \end{gathered} \]
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