Đáp án:
$\begin{array}{l}
a)y = \frac{2}{7}{x^3}\sqrt x - \frac{4}{{11}}{x^5}\sqrt x + \frac{2}{{15}}{x^7}\sqrt x \\
= \frac{2}{7}{x^{3 + \frac{1}{2}}} - \frac{4}{{11}}{x^{5 + \frac{1}{2}}} + \frac{2}{{15}}.{x^{7 + \frac{1}{2}}}\\
= \frac{2}{7}{x^{\frac{7}{2}}} - \frac{4}{{11}}.{x^{\frac{{11}}{2}}} + \frac{2}{{15}}.{x^{\frac{{15}}{2}}}\\
\Rightarrow y' = \frac{2}{7}.\frac{7}{2}.{x^{\frac{5}{2}}} - \frac{4}{{11}}.\frac{{11}}{2}.{x^{\frac{9}{2}}} + \frac{2}{{15}}.\frac{{15}}{2}.{x^{\frac{{13}}{2}}}\\
= {x^2}\sqrt x - 2{x^4}\sqrt x + {x^6}\sqrt x \\
c)y = x{e^x}\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx}}} \right)\\
y' = {e^x}.\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx}}} \right) + x.{e^x}.\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx}}} \right) + x.{e^x}.\left( {\cos x - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\\
= {e^x}.\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx}}} \right) + 2.x.{e^x}.c{\rm{osx}}\\
{\rm{ = }}{{\rm{e}}^x}.\left( {{\mathop{\rm s}\nolimits} {\rm{inx + cosx + 2x}}{\rm{.cosx}}} \right)\\
f)y = \ln \sqrt {\frac{{1 + {\mathop{\rm sinx}\nolimits} }}{{1 - {\mathop{\rm s}\nolimits} {\rm{inx}}}}} = \ln \sqrt { - 1 + \frac{2}{{1 - {\mathop{\rm s}\nolimits} {\rm{inx}}}}} \\
\Rightarrow y' = \frac{1}{2}.\frac{{ - 2\cos x}}{{{{\left( {1 - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^2}}}.\frac{1}{{\sqrt { - 1 + \frac{2}{{1 - {\mathop{\rm s}\nolimits} {\rm{inx}}}}} }}:\sqrt { - 1 + \frac{2}{{1 - {\mathop{\rm s}\nolimits} {\rm{inx}}}}} \\
= \frac{{ - c{\rm{osx}}}}{{{{\left( {1 - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^2}.\frac{{1 - {\mathop{\rm s}\nolimits} {\rm{inx}}}}{{1 + {\mathop{\rm s}\nolimits} {\rm{inx}}}}}}\\
= - \frac{{{\rm{cosx}}\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}}{{{{\left( {1 - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}^3}}}\\
= \frac{{{\rm{cosx}}\left( {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)}}{{{{\left( {{\mathop{\rm s}\nolimits} {\rm{inx - 1}}} \right)}^3}}}
\end{array}$
