Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
11,\\
y = {\cos ^2}\sqrt {2x + \dfrac{\pi }{4}} \\
\Rightarrow y' = 2.\left( {\cos \sqrt {2x + \dfrac{\pi }{4}} } \right)'.\cos \sqrt {2x + \dfrac{\pi }{4}} \\
= 2.\sqrt {2x + \dfrac{\pi }{4}} '.\left( { - \sin \sqrt {2x + \dfrac{\pi }{4}} } \right).\cos \sqrt {2x + \dfrac{\pi }{4}} \\
= - 2.\dfrac{{\left( {2x + \dfrac{\pi }{4}} \right)'}}{{2\sqrt {2x + \dfrac{\pi }{4}} }}.\sin \sqrt {2x + \dfrac{\pi }{4}} .\cos \sqrt {2x + \dfrac{\pi }{4}} \\
= - 2.\dfrac{2}{{2\sqrt {2x + \dfrac{\pi }{4}} }}.\dfrac{1}{2}\sin 2\sqrt {2x + \dfrac{\pi }{4}} \\
= - \dfrac{1}{{\sqrt {2x + \dfrac{\pi }{4}} }}.\sin \sqrt {8x + \pi } \\
= \dfrac{{ - \sin \sqrt {8x + \pi } }}{{\sqrt {2x + \dfrac{\pi }{4}} }}\\
12,\\
y = \dfrac{{1 - 3x}}{{\sqrt {4x - 1} }}\\
\Rightarrow y' = \dfrac{{\left( {1 - 3x} \right)'.\sqrt {4x - 1} - \sqrt {4x - 1} '.\left( {1 - 3x} \right)}}{{4x - 1}}\\
= \dfrac{{ - 3\sqrt {4x - 1} - \dfrac{{\left( {4x - 1} \right)'}}{{2\sqrt {4x - 1} }}.\left( {1 - 3x} \right)}}{{4x - 1}}\\
= \dfrac{{ - 3\sqrt {4x - 1} - \dfrac{2}{{\sqrt {4x - 1} }}\left( {1 - 3x} \right)}}{{4x - 1}}\\
= \dfrac{{ - 3{{\sqrt {4x - 1} }^2} - 2\left( {1 - 3x} \right)}}{{\left( {4x - 1} \right)\sqrt {4x - 1} }}\\
= \dfrac{{ - 3\left( {4x - 1} \right) - 2 + 6x}}{{\left( {4x - 1} \right)\sqrt {4x - 1} }}\\
= \dfrac{{ - 6x + 1}}{{\left( {4x - 1} \right)\sqrt {4x - 1} }}
\end{array}\)
