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: 65

Answer

\[{y^,} = \frac{{2{x^3} - \sin 2x}}{{\sqrt {{x^4} + \cos 2x} }}\]

Work Step by Step

\[\begin{gathered} y = \sqrt {{x^4} + \cos 2x} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ {y^,} = {\left( {{x^4} + \cos 2x} \right)^{1/2}} \hfill \\ \hfill \\ use\,\,the\,\,Chain\,\,rule \hfill \\ \hfill \\ {y^,} = \frac{1}{2}{\left( {{x^4} + \cos 2x} \right)^{ - 1/2}}{\left( {{x^4} + \cos 2x} \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {y^,} = \frac{{4{x^3} - \sin 2x \cdot \,{{\left( {2x} \right)}^,}}}{{2\sqrt {{x^4} + \cos 2x} }} \hfill \\ \hfill \\ {y^,} = \frac{{4{x^3} - 2\sin 2x}}{{2\sqrt {{x^4} + \cos 2x} }} \hfill \\ \hfill \\ {y^,} = \frac{{2{x^3} - \sin 2x}}{{\sqrt {{x^4} + \cos 2x} }} \hfill \\ \hfill \\ \end{gathered} \]
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