Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
13,\\
y = \left( {{x^3} + x} \right)\left( {x - 1} \right) = {x^4} - {x^3} + {x^2} - x\\
\Rightarrow y' = 4{x^3} - 3{x^2} + 2x - 1\\
14,\\
y = \dfrac{{ - 2 - 3x}}{{2x + 1}}\\
\Rightarrow y' = \dfrac{{\left( { - 2 - 3x} \right)'.\left( {2x + 1} \right) - \left( {2x + 1} \right)'.\left( { - 2 - 3x} \right)}}{{{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{ - 3\left( {2x + 1} \right) - 2.\left( { - 2 - 3x} \right)}}{{{{\left( {2x + 1} \right)}^2}}}\\
= \dfrac{{ - 6x - 3 + 4 + 6x}}{{{{\left( {2x + 1} \right)}^2}}} = \dfrac{1}{{{{\left( {2x + 1} \right)}^2}}}\\
15,\\
y = {x^2} + \sqrt x \Rightarrow y' = 2x + \dfrac{1}{{2\sqrt x }}\\
16,\\
f\left( x \right) = 2\cos x \Rightarrow f'\left( x \right) = - 2\sin x\\
17,\\
y = {\cos ^2}\left( {\dfrac{\pi }{4} - x} \right)\\
\Rightarrow y' = 2.\left[ {\cos \left( {\dfrac{\pi }{4} - x} \right)} \right]'.\cos \left( {\dfrac{\pi }{4} - x} \right)\\
= 2.\left( {\dfrac{\pi }{4} - x} \right)'.\left( { - \sin \left( {\dfrac{\pi }{4} - x} \right)} \right).\cos \left( {\dfrac{\pi }{4} - x} \right)\\
= 2.\left( { - 1} \right).\left( { - \sin \left( {\dfrac{\pi }{4} - x} \right)} \right).\cos \left( {\dfrac{\pi }{4} - x} \right)\\
= 2\sin \left( {\dfrac{\pi }{4} - x} \right).\cos \left( {\dfrac{\pi }{4} - x} \right)\\
= \sin \left( {\dfrac{\pi }{2} - 2x} \right)\\
= \cos 2x\\
18,\\
y = {\left( {{x^3} - 2{x^2}} \right)^{2020}}\\
\Rightarrow y' = 2020.\left( {{x^3} - 2{x^2}} \right)'.{\left( {{x^3} - 2{x^2}} \right)^{2019}}\\
= 2020.\left( {3{x^2} - 4x} \right){\left( {{x^3} - 2{x^2}} \right)^{2019}}
\end{array}\)
