If
f
(
x
)
=
x
+
5
x
−
3
\displaystyle {f{{\left({x}\right)}}}=\frac{{\sqrt{{{x}}}+{5}}}{{\sqrt{{{x}}}-{3}}}
f
(
x
)
=
x
−
3
x
+
5
, then:
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
=
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