Đáp án:
$\begin{array}{l}
c)y = \dfrac{2}{{{{\left( {x + 1} \right)}^4}}} = 2.{\left( {x + 1} \right)^{ - 4}}\\
\Rightarrow y' = - 4.2.{\left( {x + 1} \right)^{ - 4 - 1}}\\
= - 8.\dfrac{1}{{{{\left( {x + 1} \right)}^5}}} = \dfrac{{ - 8}}{{{{\left( {x + 1} \right)}^5}}}\\
f)y = \left( {{x^2} + 2} \right).{e^{2x}}\\
y' = \left( {{x^2} + 2} \right)'.{e^{2x}} + \left( {{x^2} + 2} \right).\left( {{e^{2x}}} \right)'\\
= 2x.{e^{2x}} + \left( {{x^2} + 2} \right).2.{e^{2x}}\\
= \left( {2{x^2} + 2x + 4} \right).{e^{2x}}\\
g)y = sin\left( {2x + \dfrac{\pi }{4}} \right) - \tan \dfrac{x}{2}\\
\Rightarrow y' = 2.\cos \left( {2x + \dfrac{\pi }{4}} \right) - \dfrac{1}{2}.\dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}}
\end{array}$
