Let
f
(
t
)
=
2
t
4
−
8
t
+
6
e
t
\displaystyle {f{{\left({t}\right)}}}={2}{t}^{{4}}-{8}{t}+{6}{e}^{{t}}
f
(
t
)
=
2
t
4
−
8
t
+
6
e
t
.
Then
f
′
(
t
)
=
\displaystyle {f}'{\left({t}\right)}=
f
′
(
t
)
=
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