Let
f
(
x
)
=
x
5
⋅
(
x
+
5
x
+
3
)
4
\displaystyle {f{{\left({x}\right)}}}={x}^{{5}}\cdot{\left(\frac{{{x}+{5}}}{{{x}+{3}}}\right)}^{{4}}
f
(
x
)
=
x
5
⋅
(
x
+
3
x
+
5
)
4
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
=
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