Đáp án:
a. \( - \dfrac{4}{3}{\left( {x + 5} \right)^{\dfrac{4}{3}}} + {\left( {x + 5} \right)^{\dfrac{1}{3}}}.\dfrac{{ - 16x + 32}}{9}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.y = \dfrac{{ - 4x + 8}}{3}.{\left( {\sqrt[3]{{{{\left( {x + 5} \right)}^2}}}} \right)^2}\\
= \dfrac{{ - 4x + 8}}{3}.\sqrt[3]{{{{\left( {x + 5} \right)}^4}}}\\
= \dfrac{{ - 4x + 8}}{3}.{\left( {x + 5} \right)^{\dfrac{4}{3}}}\\
y' = \dfrac{{ - 4.3}}{9}.{\left( {x + 5} \right)^{\dfrac{4}{3}}} + \dfrac{4}{3}{\left( {x + 5} \right)^{\dfrac{1}{3}}}.\dfrac{{ - 4x + 8}}{3}\\
= - \dfrac{4}{3}{\left( {x + 5} \right)^{\dfrac{4}{3}}} + {\left( {x + 5} \right)^{\dfrac{1}{3}}}.\dfrac{{ - 16x + 32}}{9}\\
b.y = \dfrac{{2{x^4} - 18{x^2}}}{{\sqrt {{{\left( {{x^2} - 6} \right)}^3}} }}\\
= \dfrac{{2{x^4} - 18{x^2}}}{{{{\left( {{x^2} - 6} \right)}^{\dfrac{3}{2}}}}}\\
y' = \dfrac{{\left( {4.2.{x^3} - 36x} \right){{\left( {{x^2} - 6} \right)}^{\dfrac{3}{2}}} - \dfrac{3}{2}.2x{{\left( {{x^2} - 6} \right)}^{\dfrac{1}{2}}}\left( {2{x^4} - 18{x^2}} \right)}}{{{{\left( {{x^2} - 6} \right)}^3}}}\\
= \dfrac{{\left( {8{x^3} - 36x} \right)\sqrt {{{\left( {{x^2} - 6} \right)}^3}} - 3x\left( {2{x^4} - 18{x^2}} \right)\sqrt {{x^2} - 6} }}{{{{\left( {{x^2} - 6} \right)}^3}}}
\end{array}\)
