Let
f
(
x
)
=
3
x
2
+
4
x
+
3
\displaystyle {f{{\left({x}\right)}}}=\sqrt{{{3}{x}^{{2}}+{4}{x}+{3}}}
f
(
x
)
=
3
x
2
+
4
x
+
3
(Use sqrt(N) to write
N
\displaystyle \sqrt{{{N}}}
N
)
f
′
(
x
)
=
\displaystyle {f}'{\left({x}\right)}=
f
′
(
x
)
=
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