If
f
(
x
)
=
2
−
cosh
x
8
+
cosh
x
\displaystyle {f{{\left({x}\right)}}}={\frac{{{2}-{\cosh{{x}}}}}{{{8}+{\cosh{{x}}}}}}
f
(
x
)
=
8
+
cosh
x
2
−
cosh
x
then
f
′
(
x
)
=
\displaystyle {f}'{\left({x}\right)}=
f
′
(
x
)
=
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