Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises: 80

Answer

\[f\,\left( x \right) = - \frac{1}{{2\ln 2\,\left( {x + 1} \right)}}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {\log _2}\frac{8}{{\sqrt {x + 1} }} \hfill \\ \hfill \\ Rewrite \hfill \\ \hfill \\ f\,\left( x \right) = {\log _2}8 - {\log _2}\,{\left( {x + 1} \right)^{\frac{1}{2}}} \hfill \\ \hfill \\ f\,\left( x \right) = {\log _2}8 - \frac{1}{2}{\log _2}\,\left( {x + 1} \right) \hfill \\ \hfill \\ Differentiate,{\text{ use }}\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a\left( u \right)}}{u^,} \hfill \\ \hfill \\ f\,\left( x \right) = 0 - \frac{1}{2}\,\left( {\frac{1}{{\,\left( {x + 1} \right)\ln 2}}} \right)\,\left( 1 \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ f\,\left( x \right) = - \frac{1}{{2\ln 2\,\left( {x + 1} \right)}} \hfill \\ \end{gathered} \]
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