Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
y = {\cos ^2}\sqrt {\dfrac{\pi }{4} - 2x} \\
\Rightarrow y' = 2.\left( {\cos \sqrt {\dfrac{\pi }{4} - 2x} } \right)'.\cos \sqrt {\dfrac{\pi }{4} - 2x} \\
= 2.\sqrt {\dfrac{\pi }{4} - 2x} '.\left( { - \sin \sqrt {\dfrac{\pi }{4} - 2x} } \right).\cos \sqrt {\dfrac{\pi }{4} - 2x} \\
= - \dfrac{{\left( {\dfrac{\pi }{4} - 2x} \right)'}}{{2\sqrt {\dfrac{\pi }{4} - 2x} }}.\left( {2\sin \sqrt {\dfrac{\pi }{4} - 2x} .\cos \sqrt {\dfrac{\pi }{4} - 2x} } \right)\\
= - \dfrac{{ - 2}}{{2\sqrt {\dfrac{\pi }{4} - 2x} }}.sin2\sqrt {\dfrac{\pi }{4} - 2x} \\
= \dfrac{{\sin \sqrt {\pi - 8x} }}{{\sqrt {\dfrac{\pi }{4} - 8x} }}\\
b,\\
y = \dfrac{{{{\sin }^2}x}}{{1 + \tan 2x}}\\
\Rightarrow y' = \dfrac{{\left( {{{\sin }^2}x} \right)'.\left( {1 + \tan 2x} \right) - \left( {1 + \tan 2x} \right)'.si{n^2}x}}{{{{\left( {1 + \tan 2x} \right)}^2}}}\\
= \dfrac{{2.\left( {\sin x} \right)'.\sin x.\left( {1 + \tan 2x} \right) - \dfrac{{\left( {2x} \right)'}}{{{{\cos }^2}2x}}.{{\sin }^2}x}}{{{{\left( {1 + \tan 2x} \right)}^2}}}\\
= \dfrac{{2\sin x.\cos x.\left( {1 + \tan 2x} \right) - \dfrac{{2{{\sin }^2}x}}{{{{\cos }^2}2x}}}}{{{{\left( {1 + \tan 2x} \right)}^2}}}\\
= \dfrac{{\sin 2x.{{\cos }^2}2x.\left( {1 + \tan 2x} \right) - 2{{\sin }^2}x}}{{{{\cos }^2}2x.{{\left( {1 + \tan 2x} \right)}^2}}}
\end{array}\)
