#### Answer

\[ = \frac{7}{{2\sqrt {7x - 1} }}\]

#### Work Step by Step

\[\begin{gathered}
y = \sqrt {7x - 1} \hfill \\
\hfill \\
y = f\,\left( u \right) = \sqrt u \hfill \\
\hfill \\
set\,\,u = g\,\left( x \right) = 7x - 1 \hfill \\
\hfill \\
Use\,\,the\,\,version\,\,1\,\,of\,\,the\,\,chain\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}} \hfill \\
\hfill \\
Therefore \hfill \\
\hfill \\
\,\,\frac{{dy}}{{dx}} = \frac{d}{{du}}\,\left( {\sqrt u } \right) \cdot \frac{d}{{dx}}\,\left( {7x - 1} \right) \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \frac{1}{{2\sqrt u }} \cdot 7 \hfill \\
\hfill \\
substitute\,\,\,back\,\,u = 7x - 1 \hfill \\
\hfill \\
= \frac{7}{{2\sqrt {7x - 1} }} \hfill \\
\hfill \\
\end{gathered} \]