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: 13

Answer

\[\frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} + 1} }}\]

Work Step by Step

\[\begin{gathered} y = \sqrt {{x^2} + 1} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ set\,\,y = \,{\left( {{x^2} + 1} \right)^{\frac{1}{2}}} \hfill \\ \hfill \\ Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\ \hfill \\ set\,\,\,u = {x^2} + 1 \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{1}{2}\,{\left( {{x^2} + 1} \right)^{ - \frac{1}{2}}}\,\left( {{x^2} + 1} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{x^2} + 1} }}\,\left( {2x} \right) \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} + 1} }} \hfill \\ \hfill \\ \end{gathered} \]
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