Find the first derivative of
g
(
t
)
=
ln
(
t
3
e
t
)
\displaystyle {g{{\left({t}\right)}}}={\ln{{\left({t}^{{3}}{e}^{{t}}\right)}}}
g
(
t
)
=
ln
(
t
3
e
t
)
.
g
′
(
t
)
=
\displaystyle {g}'{\left({t}\right)}=
g
′
(
t
)
=
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