a) \(y' = 2\pi {\left( {2x + 1} \right)^{\pi - 1}}\)
b) Áp dụng: \(\left( {\root n \of u } \right)' = {u \over {n\root n \of {{u^{n - 1}}} }}\)
\(y' = {{\left( {{{\ln }^3}5x} \right)'} \over {5\root 5 \of {{{\left( {{{\ln }^3}5x} \right)}^4}} }} = {{3{{\ln }^2}5x} \over {5x\root 5 \of {{{\ln }^{12}}5x} }}\)
c) Đặt \(u = {{1 + {x^3}} \over {1 - {x^3}}};\,\,y' = {{u'} \over {3\root 3 \of {{u^2}} }}\)
\(u' = {{3{x^2}\left( {1 - {x^3}} \right) - 3{x^2}\left( {1 + {x^3}} \right)} \over {{{\left( {1 - {x^3}} \right)}^2}}} = {{6{x^2}} \over {{{\left( {1 - {x^3}} \right)}^2}}}\)
Do đó: \(y' = {{2{x^2}} \over {{{\left( {1 - {x^3}} \right)}^2}}}.{1 \over {\root 3 \of {{{\left( {{{1 + {x^3}} \over {1 - {x^3}}}} \right)}^2}} }} = {{2{x^2}} \over {\root 3 \of {{{\left( {1 - {x^3}} \right)}^4}{{\left( {1 + {x^3}} \right)}^2}} }}\)
d)
\(\eqalign{
& y' = \left[ {{{\left( {{x \over b}} \right)}^a}} \right]'{\left( {{a \over x}} \right)^b} + {\left( {{x \over b}} \right)^a}\left[ {{{\left( {{a \over x}} \right)}^b}} \right]' \cr
& \,\,\,\,\,\, = {a \over b}{\left( {{x \over a}} \right)^{a - 1}}{\left( {{a \over x}} \right)^b} + {\left( {{x \over b}} \right)^a}b{\left( {{a \over x}} \right)^{b - 1}}\left( { - {a \over {{x^2}}}} \right) = {\left( {{x \over b}} \right)^a}{\left( {{a \over x}} \right)^b}{{a - b} \over x} \cr} \)
