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