## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises - Page 244: 48

#### Answer

${h^,}\,\left( z \right) = \frac{{{e^z}}}{{\,\left( {1 + {e^z}} \right)\ln 10}}$

#### Work Step by Step

$\begin{gathered} h\,\left( z \right) = \log \,\left( {1 + {e^z}} \right) \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\ {h^,}\,\left( z \right) = \,\,\left[ {\log \,\left( {1 + {e^z}} \right)} \right] \hfill \\ Use\,\,the\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\left[ {{{\log }_a}g\,\left( x \right)} \right] = \frac{1}{{\ln a}}\,\left( {\frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}}} \right)\, \hfill \\ Then \hfill \\ {h^,}\,\left( z \right) = \frac{1}{{\ln 10}}\,\left( {\frac{{\,\left( {1 + {e^z}} \right)}}{{1 + {e^z}}}} \right) \hfill \\ {h^,}\,\left( z \right) = \frac{1}{{\ln 10}}\,\left( {\frac{{{e^z}}}{{1 + {e^z}}}} \right) \hfill \\ Multiplying \hfill \\ {h^,}\,\left( z \right) = \frac{{{e^z}}}{{\,\left( {1 + {e^z}} \right)\ln 10}} \hfill \\ \end{gathered}$

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