m.n giúp e câu này vs ạ e vote 5 sao cho ạ e cảm ơn ^-^ tính đạo hàm a)y= $\frac{sin2x+cos2x}{2sin2x-cos2x}$ b)y=sin(cos3x) c)y= $\frac{sin2x+cos2x}

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
y = \frac{{\sin 2x + \cos 2x}}{{2\sin 2x – \cos 2x}}\\
 \Rightarrow y’ = \frac{{\left( {\sin 2x + \cos 2x} \right)’.\left( {2\sin 2x – \cos 2x} \right) – \left( {2\sin 2x – \cos 2x} \right)’.\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{\left( {2.cos2x + 2.\left( { – \sin 2x} \right)} \right)\left( {2\sin 2x – \cos 2x} \right) – \left( {2.2\cos 2x – 2.\left( { – \sin 2x} \right)} \right).\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{\left( {2\cos 2x – 2\sin 2x} \right)\left( {2\sin 2x – \cos 2x} \right) – \left( {4\cos 2x + 2\sin 2x} \right)\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 2{{\cos }^2}2x + 6\sin 2x.\cos 2x – 4{{\sin }^2}2x – 4{{\cos }^2}2x – 6\sin 2x.\cos 2x – 2{{\sin }^2}2x}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 6{{\cos }^2}2x – 6{{\sin }^2}2x}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 6}}{{{{\left( {2\sin 2x – \cos 2x} \right)}^2}}}\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}2x + {{\cos }^2}2x = 1} \right)\\
b,\\
y = \sin \left( {\cos 3x} \right)\\
y’ = \left( {\cos 3x} \right)’.cos\left( {\cos 3x} \right)\\
 = \left( {3x} \right)’.\left( { – \sin 3x} \right).\cos \left( {\cos \left( {3x} \right)} \right)\\
 =  – 3.\sin 3x.\cos \left( {\cos 3x} \right)\\
c,\\
y = \frac{{\sin 2x + \cos 2x}}{{\sin 2x – \cos 2x}}\\
 \Rightarrow y’ = \frac{{\left( {\sin 2x + \cos 2x} \right)’.\left( {\sin 2x – \cos 2x} \right) – \left( {\sin 2x – \cos 2x} \right)’.\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{\left( {2.cos2x + 2.\left( { – \sin 2x} \right)} \right)\left( {\sin 2x – \cos 2x} \right) – \left( {2\cos 2x – 2.\left( { – \sin 2x} \right)} \right).\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{\left( {2\cos 2x – 2\sin 2x} \right)\left( {\sin 2x – \cos 2x} \right) – \left( {2\cos 2x + 2\sin 2x} \right)\left( {\sin 2x + \cos 2x} \right)}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 2{{\cos }^2}2x + 4\sin 2x.\cos 2x – 2{{\sin }^2}2x – 2{{\cos }^2}2x – 4\sin 2x.\cos 2x – 2{{\sin }^2}2x}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 4{{\cos }^2}2x – 4{{\sin }^2}2x}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}\\
 = \frac{{ – 4}}{{{{\left( {\sin 2x – \cos 2x} \right)}^2}}}
\end{array}\)