If
f
(
x
)
=
2
sin
−
1
(
x
4
)
\displaystyle {f{{\left({x}\right)}}}={2}{{\sin}^{{-{{1}}}}{\left({x}^{{{4}}}\right)}}
f
(
x
)
=
2
sin
−
1
(
x
4
)
, then
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
=
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