Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y = \frac{{\sin x + \cos x}}{{\cos x + 1}}\\
\Rightarrow y' = \frac{{\left( {\sin x + \cos x} \right)'.\left( {\cos x + 1} \right) - \left( {\cos x + 1} \right)'.\left( {\sin x + \cos x} \right)}}{{{{\left( {\cos x + 1} \right)}^2}}}\\
= \frac{{\left( {\cos x - \sin x} \right).\left( {\cos x + 1} \right) + \sin x.\left( {\sin x + \cos x} \right)}}{{{{\left( {\cos x + 1} \right)}^2}}}\\
= \frac{{{{\cos }^2}x + \cos x - \sin x\cos x - \sin x + {{\sin }^2}x + \sin x\cos x}}{{{{\left( {\cos x + 1} \right)}^2}}}\\
= \frac{{1 + \cos x - \sin x}}{{{{\left( {\cos x + 1} \right)}^2}}}\\
b,\\
y = 3{\sin ^3}x\\
\Rightarrow y' = 3.3.\left( {\sin x} \right)'.{\sin ^2}x = 9.\cos x.{\sin ^2}x\\
c,\\
y = \frac{{2{{\cot }^9}\left( {2x - 11} \right)}}{3}\\
\Rightarrow y' = \frac{2}{3}.9.\left( {2x - 11} \right)'.\left( {\cot \left( {2x - 11} \right)} \right)'.co{t^8}\left( {2x - 11} \right)\\
= 6.2.\frac{{ - 1}}{{{{\sin }^2}\left( {2x - 11} \right)}}.{\cot ^8}\left( {2x - 11} \right)\\
= \frac{{ - 12{{\cot }^8}\left( {2x - 11} \right)}}{{{{\sin }^2}\left( {2x - 11} \right)}}
\end{array}\)
