Let
f
(
x
)
=
−
5
ln
(
sin
(
x
)
)
\displaystyle {f{{\left({x}\right)}}}=-{5}{\ln{{\left({\sin{{\left({x}\right)}}}\right)}}}
f
(
x
)
=
−
5
ln
(
sin
(
x
)
)
f
′
′
(
x
)
=
\displaystyle {f}{''}{\left({x}\right)}=
f
′′
(
x
)
=
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