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: 26

Answer

\[\frac{{dy}}{{dx}} = {4^{ - x}}\cos x - \ln 4\sin x \cdot {4^{ - x}}\]

Work Step by Step

\[\begin{gathered} y = {4^{ - x}}\sin x \hfill \\ \hfill \\ Use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {{4^{ - x}}} \right)\,{\left( {\sin x} \right)^,} + \,\left( {\sin x} \right)\,{\left( {{4^{ - x}}} \right)^,} \hfill \\ \hfill \\ Use\,\,the\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{a^u}} \right] = {a^u}\ln a \cdot {u^,} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {4^{ - x}}\cos x + \sin x\,\left( {{4^{ - x}}\ln 4\,\left( { - 1} \right)} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {4^{ - x}}\cos x - \ln 4\sin x \cdot {4^{ - x}} \hfill \\ \end{gathered} \]
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