Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
y = {x^2} - 5x - 6\\
 \Rightarrow DKXD:\,\,\,\,\,D = R\\
b,\\
y = {\left( {{x^2} - 4} \right)^{ - 1}}\\
DKXD:\,\,\,\,\,\,{x^2} - 4 \ne 0 \Leftrightarrow {x^2} \ne 4 \Leftrightarrow x \ne  \pm 2\\
2,\\
a,\\
y = \sqrt[3]{{2x}} - 1 = {\left( {2x} \right)^{\dfrac{1}{3}}} - 1\\
 \Rightarrow y' = \dfrac{1}{3}.\left( {2x} \right)'.{\left( {2x} \right)^{\dfrac{1}{3} - 1}} = \dfrac{1}{3}.2.{\left( {2x} \right)^{ - \dfrac{2}{3}}} = \dfrac{2}{{3.{{\left( {2x} \right)}^{\dfrac{2}{3}}}}} = \dfrac{2}{{3.\sqrt[3]{{{{\left( {2x} \right)}^2}}}}} = \dfrac{2}{{3\sqrt[3]{{4{x^2}}}}}\\
b,\\
y = 3x + 1 \Rightarrow y' = 3\\
c,\\
y = \dfrac{{\sqrt[5]{x} + 1}}{{\sqrt x  - 1}} = \dfrac{{{x^{\dfrac{1}{5}}} + 1}}{{{x^{\dfrac{1}{2}}} - 1}}\\
 \Rightarrow y' = \dfrac{{\left( {{x^{\dfrac{1}{5}}} + 1} \right)'.\left( {{x^{\dfrac{1}{2}}} - 1} \right) - \left( {{x^{\dfrac{1}{2}}} - 1} \right)'.\left( {{x^{\dfrac{1}{5}}} + 1} \right)}}{{{{\left( {{x^{\dfrac{1}{2}}} - 1} \right)}^2}}}\\
 = \dfrac{{\dfrac{1}{5}.{x^{\dfrac{1}{5} - 1}}.\left( {{x^{\dfrac{1}{2}}} - 1} \right) - \dfrac{1}{2}.{x^{\dfrac{1}{2} - 1}}.\left( {{x^{\dfrac{1}{5}}} + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{{\dfrac{1}{5}{x^{\dfrac{{ - 4}}{5}}}.\left( {{x^{\dfrac{1}{2}}} - 1} \right) - \dfrac{1}{2}{x^{\dfrac{{ - 1}}{2}}}.\left( {{x^{\dfrac{1}{5}}} + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{{\dfrac{1}{5}.{x^{ - \dfrac{3}{{10}}}} - \dfrac{1}{5}{x^{ - \dfrac{1}{5}}} - \dfrac{1}{2}{x^{ - \dfrac{3}{{10}}}} - \dfrac{1}{2}{x^{ - \dfrac{1}{2}}}}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{{ - \dfrac{3}{{10}}{x^{ - \dfrac{3}{{10}}}} - \dfrac{1}{5}{x^{ - \dfrac{1}{5}}} - \dfrac{1}{2}{x^{ - \dfrac{1}{2}}}}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}\\
 = \dfrac{{ - 3\sqrt[{10}]{{{x^{ - 3}}}} - 2.\sqrt[5]{{{x^{ - 1}}}} - 5.\sqrt[2]{{{x^{ - 1}}}}}}{{10.{{\left( {\sqrt x  - 1} \right)}^2}}}
\end{array}\)