# Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 53

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

#### Work Step by Step

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

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.