Đáp án:
$\begin{array}{l}
a)y' = \dfrac{{2x - 5}}{{2\sqrt {{x^2} - 5x + 2} }}\\
b) - 12{x^3} + 4x\\
c)y' = \dfrac{2}{3}.6{x^5} + 4.2x = 4{x^5} + 8x\\
d)y' = \dfrac{{\left( {1 - 4x} \right)\sin \sqrt {2{x^2} - x + 7} }}{{2\sqrt {2{x^2} - x + 7} }}
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)y = \sqrt {{x^2} - 5x + 2} \\
\Rightarrow y' = \dfrac{{2x - 5}}{{2\sqrt {{x^2} - 5x + 2} }}\\
b)y = \left( {{x^2} + 1} \right)\left( {5 - 3{x^2}} \right)\\
\Rightarrow y' = 2x\left( {5 - 3{x^2}} \right) + \left( {{x^2} + 1} \right).\left( { - 6x} \right)\\
= - 12{x^3} + 4x\\
c)y = \dfrac{2}{3}{x^6} + 4{x^2} + \sqrt {2021} \\
\Rightarrow y' = \dfrac{2}{3}.6{x^5} + 4.2x = 4{x^5} + 8x\\
d)y = \cos \sqrt {2{x^2} - x + 7} \\
\Rightarrow y' = - \sin \sqrt {2{x^2} - x + 7} .\left( {\dfrac{{4x - 1}}{{2\sqrt {2{x^2} - x + 7} }}} \right)\\
= \dfrac{{\left( {1 - 4x} \right)\sin \sqrt {2{x^2} - x + 7} }}{{2\sqrt {2{x^2} - x + 7} }}
\end{array}$
