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