Answer
\[ \frac{d}{dx}(|sin(x)| = \begin{cases}
cos(x) & x >0 \\
-cos(x) & x<0 \\
\end{cases}
\]
Work Step by Step
We are given that
\[\frac{d}{dx}|x|= \begin{cases}
1 & x >0 \\
-1& x<0 \\
\end{cases}
\]
We apply the chain rule:
\[ \frac{d}{dx}(|sin(x)| = \begin{cases}
1 * (\frac{d}{dx} sin(x)) & x >0 \\
-1 * (\frac{d}{dx} sin(x)) & x<0 \\
\end{cases}
\]
\[ \frac{d}{dx}(|sin(x)| = \begin{cases}
cos(x) & x >0 \\
-cos(x) & x<0 \\
\end{cases}
\]
