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: 27

Answer

\[\frac{{dy}}{{dx}} = {x^3}\ln 3 \cdot {3^x} + {3^{x + 1}}{x^2}\]

Work Step by Step

\[\begin{gathered} y = {x^3} \cdot {3^x} \hfill \\ \hfill \\ Use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {{x^3}} \right)\,{\left( {{3^x}} \right)^,} + \,\left( {{3^x}} \right)\,{\left( {{x^3}} \right)^,} \hfill \\ \hfill \\ Use\,\,the\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{a^u}} \right] = {a^u}\ln a \cdot {u^,} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \,\left( {{x^3}} \right)\,\left( {{3^x}\ln x} \right) + {3^x}\,\left( {3{x^2}} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {x^3}\ln 3 \cdot {3^x} + {3^{x + 1}}{x^2} \hfill \\ \end{gathered} \]
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.