Đáp án:
$\begin{array}{l}
a)f\left( x \right) = \left( {{x^2} - 3x + 1} \right).\left( {1 - 3x} \right)\\
f'\left( x \right) = \left( {2x - 3} \right).\left( {1 - 3x} \right) - 3.\left( {{x^2} - 3x + 1} \right)\\
= 2x - 6{x^2} - 3 + 9x - 3{x^2} + 9x - 3\\
= - 9{x^2} + 20x - 6\\
b)f\left( x \right) = {\left( {\sin \left( {\tan \left( {{x^4} + 1} \right)} \right)} \right)^2}\\
\Leftrightarrow f'\left( x \right) = 2.\left[ {\sin \left( {\tan \left( {{x^4} + 1} \right)} \right)} \right]'.\sin \left( {\tan \left( {{x^4} + 1} \right)} \right)\\
= 2.\left( {\tan \left( {{x^4} + 1} \right)} \right)'.\cos \left( {\tan \left( {{x^4} + 1} \right)} \right).\sin \left( {\tan \left( {{x^4} + 1} \right)} \right)\\
= 2.\left( {{x^4} + 1} \right)'.\dfrac{1}{{{{\cos }^2}\left( {{x^4} + 1} \right)}}.\cos \left( {\tan \left( {{x^4} + 1} \right)} \right).\sin \left( {\tan \left( {{x^4} + 1} \right)} \right)\\
= 2.4{x^3}.\dfrac{1}{{{{\cos }^2}\left( {{x^4} + 1} \right)}}.\dfrac{1}{2}.\sin \left( {2\tan \left( {{x^4} + 1} \right)} \right)\\
= \dfrac{{4{x^3}.\sin \left( {2\tan \left( {{x^4} + 1} \right)} \right)}}{{{{\cos }^2}\left( {{x^4} + 1} \right)}}
\end{array}$
