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: 53

Answer

\[{f^,}\,\left( x \right) = \frac{{2\sqrt x + 1}}{{4\sqrt x \sqrt {x + \sqrt x } }}\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = \sqrt {x + \sqrt x } \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ f\,\left( x \right) = {\left( {x + {x^{1/2}}} \right)^{1/2}} \hfill \\ \hfill \\ use\,\,the\,\,Chain\,\,rule \hfill \\ \hfill \\ \frac{{dy}}{{dx}} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ f\,'\left( x \right) = \frac{1}{2}{\left( {x + {x^{1/2}}} \right)^{ - 1/2}}\left( {1 + \frac{1}{{2\sqrt x }}} \right) \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{1 + \frac{1}{{2\sqrt x }}}}{{2\sqrt {x + \sqrt x } }} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ {f^,}\,\left( x \right) = \frac{{2\sqrt x + 1}}{{4\sqrt x \sqrt {x + \sqrt x } }} \hfill \\ \end{gathered} \]
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