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: 46

Answer

\[\frac{{dy}}{{dx}} = \frac{{18x - 21}}{{2\sqrt {9{x^2} - 21x + 16} }}\]

Work Step by Step

\[\begin{gathered} y = \sqrt {\,{{\left( {3x - 4} \right)}^2} + 3x} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ y = {\left( {\,{{\left( {3x - 4} \right)}^2} + 3x} \right)^{1/2}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,2\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \,\,y = {u^n} \to \frac{{dy}}{{dx}} = n{u^{n - 1}}{u^,} \hfill \\ \hfill \\ Therefore \hfill \\ \hfill \\ y = \frac{1}{2}{\left( {\,{{\left( {3x - 4} \right)}^2} + 3x} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {\,{{\left( {3x - 4} \right)}^2} + 3x} \right] \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{2\,\left( {3x - 4} \right)\,\left( 3 \right) + 3}}{{2\sqrt {\,{{\left( {3x - 4} \right)}^2} + 3x} }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{18x - 24 + 3}}{{2\sqrt {9{x^2} - 24x + 16 + 3x} }} \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{18x - 21}}{{2\sqrt {9{x^2} - 21x + 16} }} \hfill \\ \hfill \\ \end{gathered} \]
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