## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# 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

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

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