Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 62

Answer

\[ = \frac{{30\,{{\left( {3x} \right)}^4}}}{{\,{{\left( {4x + 2} \right)}^6}}}\]

Work Step by Step

\[\begin{gathered} y = \,{\left( {\frac{{3x}}{{4x + 2}}} \right)^5} \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ {y^,} = \,{\left( {\frac{{3x}}{{4x + 2}}} \right)^5} \hfill \\ \hfill \\ use\,\,the\,Chain\,\,rule \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^,} \hfill \\ \hfill \\ Quotient\,\,rule \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\frac{{3\,\left( {4x + 2} \right) - 4\,\left( {3x} \right)}}{{\,{{\left( {4x + 2} \right)}^2}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 5\,{\left( {\frac{{3x}}{{4x + 2}}} \right)^4} \cdot \,\,\frac{6}{{\,{{\left( {4x + 2} \right)}^2}}} \hfill \\ \hfill \\ = \frac{{30\,{{\left( {3x} \right)}^4}}}{{\,{{\left( {4x + 2} \right)}^6}}} \hfill \\ \end{gathered} \]
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.