Given
g
(
2
)
=
1
\displaystyle {g{{\left({2}\right)}}}={1}
g
(
2
)
=
1
,
g
′
(
2
)
=
7
\displaystyle {g}'{\left({2}\right)}={7}
g
′
(
2
)
=
7
and
f
(
x
)
=
g
(
x
)
\displaystyle {f{{\left({x}\right)}}}=\sqrt{{{g{{\left({x}\right)}}}}}
f
(
x
)
=
g
(
x
)
, find
f
′
(
2
)
\displaystyle {f}'{\left({2}\right)}
f
′
(
2
)
.
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