Đáp án:
\(f)y'' = 18{x^4} + 24{x^2} + 6\)
Giải thích các bước giải:
\(\begin{array}{l}
a)y = 5{x^4} - 2{x^3} + 5{x^2} - 4x + 7\\
y' = 20{x^3} - 6{x^2} + 10x - 4\\
y'' = 60{x^2} - 12x + 10\\
c)y = \sqrt {2x - {x^2}} \\
y' = \left( {2 - 2x} \right).\dfrac{1}{{2\sqrt {2x - {x^2}} }}\\
= \dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\
y'' = \dfrac{{ - \sqrt {2x - {x^2}} - \left( {2 - 2x} \right).\dfrac{1}{{2\sqrt {2x - {x^2}} }}.\left( {1 - x} \right)}}{{2x - {x^2}}}\\
= \dfrac{{ - \sqrt {2x - {x^2}} - \dfrac{{{{\left( {1 - x} \right)}^2}}}{{\sqrt {2x - {x^2}} }}}}{{2x - {x^2}}}\\
= \dfrac{{ - 2x + {x^2} - 1 + 2x - {x^2}}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }}\\
= \dfrac{{ - 1}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }}\\
f)y = {\left( {{x^2} + 1} \right)^3}\\
y' = 3{\left( {{x^2} + 1} \right)^2}.2x\\
= 6x{\left( {{x^2} + 1} \right)^2}\\
y'' = 6{\left( {{x^2} + 1} \right)^2} + 2x.\left( {{x^2} + 1} \right).6x\\
= 6\left( {{x^4} + 2{x^2} + 1} \right) + 12{x^4} + 12{x^2}\\
= 18{x^4} + 24{x^2} + 6
\end{array}\)
