If
f
(
x
)
=
coth
−
1
x
2
+
8
\displaystyle {f{{\left({x}\right)}}}=\text{coth}^{{-{1}}}\sqrt{{{x}^{{2}}+{8}}}
f
(
x
)
=
coth
−
1
x
2
+
8
then
f
′
(
x
)
=
\displaystyle {f}'{\left({x}\right)}=
f
′
(
x
)
=
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