Giải thích các bước giải:
$a.y=\sqrt[3]{x^3-3x+1}$
$\to y'=(x^3-3x+1)'.\dfrac{1}{3(x^3-3x+1)^{\frac{2}{3}}}$
$\to y'=\dfrac{x^2-1}{\left(x^3-3x+1\right)^{\frac{2}{3}}}$
$b.y=\sqrt[3]{\dfrac{2x-1}{(x+3)^2}}$
$\to y'=(\sqrt[3]{\dfrac{2x-1}{(x+3)^2}})'.(\dfrac{2x-1}{(x+3)^2})'$
$\to y'=\dfrac{1}{3\left(\frac{2x-1}{\left(x+3\right)^2}\right)^{\frac{2}{3}}}\cdot \dfrac{2\left(-x+4\right)}{\left(x+3\right)^3}$
$\to y'=\dfrac{2\left(\left(x+3\right)^2\right)^{\frac{2}{3}}\left(-x+4\right)}{3\left(x+3\right)^3\left(2x-1\right)^{\frac{2}{3}}}$
