Đáp án:
Giải thích các bước giải:
$a) y'=(tan\frac{x}{2}+cot\frac{x}{2})'=\frac{(\frac{x}{2})'}{cos^2(\frac{x}{2})}=\frac{1/2}{cos^2(\frac{x}{2})}\\
b) y'=(tanx-\frac{1}{3}tan^3x+\frac{1}{5}tan^5x)'=\frac{1}{cos^2x}-tan^2x.\frac{1}{cos^2x}+tan^4x.\frac{1}{cos^2x}\\
c) y'=((2-x)^2cosx+2xsinx)'=2.(-1).(2-x).cosx+(2-x)^2(-sinx)+2.sinx+2xcosx=(-4+2x)cosx-(2-x)^2sinx+2sinx+2xcosx=-4cosx+2xcosx-(2-x)^2sinx+2sinx+2xcosx=-4cosx+4xcosx-(2-x)^2sinx+2sinx\\
d) y'=(sin(cos^2x)+cos(sin^2x))'=(cos^2x)'.cos(cos^2x)-(sin^2x)'.sin(sin^2x)=2.(-sinx).cosx.cos(cos^2x)-2.sinx.cosx.sin(sin^2x)\\
e) y'=(sin3x-3cos\frac{x}{5}+tan\sqrt{x})'=3.cos3x-3.(-\frac{1}{5}).sin\frac{x}{5}+\frac{1}{2\sqrt{x}}.\frac{1}{cos^2\sqrt{x}}$
