Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 25

Answer

\[{y^,} = \frac{{{e^{{x^2}}}}}{x} + 2x{e^{{x^2}}}\ln x\]

Work Step by Step

\[\begin{gathered} y = {e^{{x^2}}}\ln x \hfill \\ Differentiate \hfill \\ {y^,} = \,\,\left[ {{e^{{x^2}}}\ln x} \right] \hfill \\ Use\,\,the\,\,product\,\,rule\, \hfill \\ {y^,} = {e^{{x^2}}}\,{\left( {\ln x} \right)^,} + \ln x\,\left( {{e^{{x^2}}}} \right) \hfill \\ Use\,\,the\,\,\,formula \hfill \\ \frac{d}{{dx}}\,\,\,\,\left[ {\ln x} \right] = \frac{1}{x} \hfill \\ \frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {y^,} = {e^{{x^2}}}\,\left( {\frac{1}{x}} \right) + \ln x\,\left( {2x{e^{{x^2}}}} \right) \hfill \\ Multipliying \hfill \\ {y^,} = \frac{{{e^{{x^2}}}}}{x} + 2x{e^{{x^2}}}\ln x \hfill \\ \end{gathered} \]
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.