Answer
\[\frac{{dy}}{{dx}} = {4^{ - x}}\cos x - \ln 4\sin x \cdot {4^{ - x}}\]
Work Step by Step
\[\begin{gathered}
y = {4^{ - x}}\sin x \hfill \\
\hfill \\
Use\,\,the\,\,product\,\,rule \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \,\left( {{4^{ - x}}} \right)\,{\left( {\sin x} \right)^,} + \,\left( {\sin x} \right)\,{\left( {{4^{ - x}}} \right)^,} \hfill \\
\hfill \\
Use\,\,the\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{a^u}} \right] = {a^u}\ln a \cdot {u^,} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {4^{ - x}}\cos x + \sin x\,\left( {{4^{ - x}}\ln 4\,\left( { - 1} \right)} \right) \hfill \\
\hfill \\
multiply \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = {4^{ - x}}\cos x - \ln 4\sin x \cdot {4^{ - x}} \hfill \\
\end{gathered} \]
