If
f
(
x
)
=
4
−
x
2
3
+
x
2
\displaystyle {f{{\left({x}\right)}}}=\frac{{{4}-{x}^{{2}}}}{{{3}+{x}^{{2}}}}
f
(
x
)
=
3
+
x
2
4
−
x
2
, find:
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
=
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