Đáp án:
\(1)y' = - 28{x^6} + 6x - 4\)
Giải thích các bước giải:
\(\begin{array}{l}
1)y' = - 28{x^6} + 6x - 4\\
2)y' = 2x.\left( { - \dfrac{1}{{{x^4}}}} \right) + \dfrac{1}{{{x^2}}}\\
= - \dfrac{2}{{{x^3}}} + \dfrac{1}{{{x^2}}}\\
3)y' = \dfrac{{ - 2x.4}}{{{{\left( {{x^2} + 1} \right)}^2}}} + 3 = - \dfrac{{8x}}{{{{\left( {{x^2} + 1} \right)}^2}}} + 3\\
4)y = \left( {{x^{ - \dfrac{1}{2}}}} \right) + \sqrt x - 4{x^3}\\
y' = - \dfrac{1}{2}.{x^{ - \dfrac{3}{2}}} + \dfrac{1}{{2\sqrt x }} - 12{x^2}\\
= - \dfrac{1}{{2x\sqrt x }} + \dfrac{1}{{2\sqrt x }} - 12{x^2}
\end{array}\)
