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: 30

Answer

\[s\,\left( t \right) = \frac{{ - {e^{ - t}} + \frac{1}{t}}}{{2\,{{\left( {{e^{ - t}} + \ln 2t} \right)}^{1/2}}}}\]

Work Step by Step

\[\begin{gathered} s\,\left( t \right) = \sqrt {{e^{ - t}} + \ln 2t} \hfill \\ \operatorname{Re} write\,\,the\,\,function \hfill \\ s\,\left( t \right) = \,{\left( {{e^{ - t}} + \ln 2t} \right)^{1/2}} \hfill \\ Use\,\,the\,\,chain\,\,rule \hfill \\ s\,\left( t \right) = \frac{1}{2}\,{\left( {{e^{ - t}} + \ln 2t} \right)^{1/2 - 1}}\,{\left( {{e^{ - t}} + \ln 2t} \right)^,} \hfill \\ Use\,\,the\,\,formulas \hfill \\ \frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ \frac{d}{{dx}}\,\,\left[ {\ln g\,\left( x \right)} \right] = \frac{{{g^,}\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\ Then \hfill \\ s\,\left( t \right) = \frac{1}{2}\,{\left( {{e^{ - t}} + \ln 2t} \right)^{ - 1/2}}\,\left( { - {e^{ - t}} + \frac{1}{t}} \right) \hfill \\ s\,\left( t \right) = \frac{{ - {e^{ - t}} + \frac{1}{t}}}{{2\,{{\left( {{e^{ - t}} + \ln 2t} \right)}^{1/2}}}} \hfill \\ \end{gathered} \]
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.