a) Ta có
$y' = [\tan(x+1) + cot(1-x)]'$
$= \dfrac{1}{\cos^2(x+1)} - \dfrac{-1}{\sin^2(1-x)}$
$= \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$
Vậy
$y' = \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$
b) Ta có
$y' = 2\cos x (-\sin x) + 2\cos(2x) [-2\sin(2x)] + 2\cos(3x) [-3 \sin(3x)]$
$= -2\sin x \cos x - 4\sin(2x) \cos(2x) - 6 \sin(3x) \cos(3x)$
$= -\sin(2x) - 2\sin(4x) - 3 \sin(6x)$
Vậy
$y'= -\sin(2x) - 2\sin(4x) - 3 \sin(6x)$.
