#### Answer

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

#### Work Step by Step

\[\begin{gathered}
f\,\left( x \right) = {\sin ^{ - 1}}\,\left( {{e^{ - 2x}}} \right) \hfill \\
\hfill \\
use\,\,the\,\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{{\sin }^{ - 1}}\,u} \right] = \frac{{{u^,}}}{{\sqrt {1 - {u^2}} }} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
{f^,}\,\left( x \right) = \frac{{ - 2{e^{ - 2x}}}}{{\sqrt {1 - {e^{ - 4x}}} }} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{f^,}\,\left( x \right) = - \frac{{2{e^{ - 2x}}}}{{\sqrt {1 - {e^{ - 4x}}} }} \hfill \\
\end{gathered} \]