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: 17

Answer

\[ = \frac{{{x^2} + 1}}{x} + 2x\ln x\]

Work Step by Step

\[\begin{gathered} \frac{d}{{dx}}\,\,\left[ {\,\left( {{x^2} + 1} \right)\ln x} \right] \hfill \\ \hfill \\ Use\,\,product\,\,rule \hfill \\ \hfill \\ = \,\left( {{x^2} + 1} \right)\frac{d}{{dx}}\,\,\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\,\left( {{x^2} + 1} \right) \hfill \\ \hfill \\ Differentiate\,\,use\,\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {\ln u} \right] = \frac{{{u^,}}}{u} \hfill \\ \hfill \\ = \,\left( {{x^2} + 1} \right)\,\left( {\frac{1}{x}} \right) + \ln x\,\left( {2x} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{{x^2} + 1}}{x} + 2x\ln x \hfill \\ \end{gathered} \]
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