Bài tập 5.2 trang 198 SBT Toán 11

Với Δx là số gia của đối số tại x₀
Ta có

\(\begin{array}{l}
\Delta y = \sqrt[3]{{{x_0} + \Delta x - 1}} - \sqrt[3]{{{x_0} - 1}}\\
 = \frac{{{x_0} + \Delta x - 1 - \left( {{x_0} - 1} \right)}}{{\sqrt[3]{{{{\left( {{x_0} + \Delta x - 1} \right)}^2}}} + \sqrt[3]{{\left( {{x_0} + \Delta x - 1} \right)\left( {{x_0} - 1} \right)}} + \sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}\\
 = \frac{{\Delta x}}{{\sqrt[3]{{{{\left( {{x_0} + \Delta x - 1} \right)}^2}}} + \sqrt[3]{{\left( {{x_0} + \Delta x - 1} \right)\left( {{x_0} - 1} \right)}} + \sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}
\end{array}\)
Suy ra 

\(\begin{array}{l}
\frac{{\Delta y}}{{\Delta x}} = \frac{{\frac{{\Delta x}}{{\sqrt[3]{{{{\left( {{x_0} + \Delta x - 1} \right)}^2}}} + \sqrt[3]{{\left( {{x_0} + \Delta x - 1} \right)\left( {{x_0} - 1} \right)}} + \sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}}}{{\Delta x}}\\
 = \frac{1}{{\sqrt[3]{{{{\left( {{x_0} + \Delta x - 1} \right)}^2}}} + \sqrt[3]{{\left( {{x_0} + \Delta x - 1} \right)\left( {{x_0} - 1} \right)}} + \sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}
\end{array}\)
Vậy

\(\begin{array}{l}
y'\left( {{x_0}} \right) = \mathop {\lim }\limits_{{\rm{\Delta }}x \to 0} \frac{1}{{\sqrt[3]{{{{\left( {{x_0} + {\rm{\Delta }}x - 1} \right)}^2}}} + \sqrt[3]{{\left( {{x_0} + {\rm{\Delta }}x - 1} \right)\left( {{x_0} - 1} \right)}} + \sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}\\
 = \frac{1}{{3\sqrt[3]{{{{\left( {{x_0} - 1} \right)}^2}}}}}
\end{array}\)

Vậy \(y'\left( 0 \right) = \frac{1}{{3\sqrt[3]{{{{\left( {0 - 1} \right)}^2}}}}} = \frac{1}{3}\), không tồn tại y′(1).