Let
f
(
x
)
=
(
cos
(
2
x
−
8
)
)
(
sin
(
−
2
x
+
4
)
)
\displaystyle {f{{\left({x}\right)}}}={\left({\cos{{\left({2}{x}-{8}\right)}}}\right)}{\left({\sin{{\left(-{2}{x}+{4}\right)}}}\right)}
f
(
x
)
=
(
cos
(
2
x
−
8
)
)
(
sin
(
−
2
x
+
4
)
)
.
Then
f
′
(
x
)
=
\displaystyle {f}'{\left({x}\right)}=
f
′
(
x
)
=
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