If
f
(
x
)
=
6
tan
−
1
(
4
sin
(
4
x
)
)
\displaystyle {f{{\left({x}\right)}}}={6}{{\tan}^{{-{1}}}{\left({4}{\sin{{\left({4}{x}\right)}}}\right)}}
f
(
x
)
=
6
tan
−
1
(
4
sin
(
4
x
)
)
,
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
=
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