\(\begin{array}{l}
1)\\
a)\quad y = 3\cos x + \sin2x\\
\to y' = 3(\cos x)' + (\sin2x)'\\
\to y' = -3\sin x + (2x)'\cos2x\\
\to y' = -3\sin x + 2\cos2x\\
b)\quad y = \cos3x - 4\sin4x\\
\to y' = (\cos3x)' - 4(\sin4x)'\\
\to y' = (3x)'(-\sin3x) - 4.(4x)'\cos4x\\
\to y' = -3\sin3x - 14\cos4x\\
c)\quad y = \sin^2x\\
\to y' = (\sin^2x)'\\
\to y'= 2\sin x.(\sin x)'\\
\to y' = 2\sin x.\cos x\\
\to y' = \sin2x\\
d)\quad y = \cos^3x\\
\to y' = (\cos^3x)'\\
\to y' = 3\cos^2x.(\cos x)'\\
\to y' = -3\cos^2x.\sin x\\
2)\\
a)\quad y = 3\sin^2x - 2\cos^3x\\
\to y' = 3(\sin^2x)' - 2(\cos^3x)'\\
\to y' = 3.2\sin x.(\sin x)' - 2.3\cos^2x.(\cos x)'\\
\to y' = 6\sin x.\cos x + 6\cos^2x.\sin x\\
b)\quad y = \cos x.\sin^2x\\
\to y' = (\cos x)'.\sin^2x + \cos x.(\sin^2x)'\\
\to y' = -\sin x.\sin^2x + \cos x.2\sin x.(\sin x)'\\
\to y' = - \sin^3x + 2\sin x.\cos^2x\\
c)\quad y = \cos3x.\cos2x\\
\to y' = (\cos3x)'.\cos2x + \cos3x.(\cos2x)\\
\to y' = (3x)'.(-\sin3x).\cos2x + \cos3x.(2x)'.(-\sin2x)\\
\to y' = -3\sin3x.\cos2x - 2\sin2x.\cos3x\\
d)\quad y = (1+\cot x)^2\\
\to y'= 2(1+\cot x).(1 + \cot x)'\\
\to y' = 2(1 + \cot x).(\cot x)'\\
\to y' = -\dfrac{2(1+\cot x)}{\sin^2x}\\
3)\\
a)\quad y = \cos x(1 + \sin x)\\
\to y' = (\cos x)'.(1+\sin x) + \cos x.(1+\sin x)'\\
\to y' = -\sin x(1+\sin x) + \cos x.(\sin x)'\\
\to y' = -\sin x - sin^2x + \cos^2x\\
\to y' = \cos2x - \sin x\\
b)\quad y = \sin x(2-\cos x)\\
\to y' = (\sin x)'.(2-\cos x) + \sin x.(2-\cos x)'\\
\to y' = \cos x(2-\cos x) + \sin x.(-\cos x)'\\
\to y' = 2\cos x - \cos^2x + \sin^2x\\
\to y' = 2\cos x - \cos2x\\
c)\quad y = (\sin x +\cos x)(2\sin x - 3\cos x)\\
\to y' = (\sin x + \cos x)'(2\sin x - 3\cos x) + (\sin x + \cos x)(2\sin x - 3\cos x)'\\
\to y' = (\cos x - \sin x)(2\sin x - 3\cos x) + (\sin x + \cos x)(2\cos x + 3\sin x)\\
\to y' = \sin^2x - \cos^2x +10\sin x\cos x\\
\to y' = 5\sin2x - \cos2x\\
4)\\
a)\quad y = \dfrac{x}{1-\cos x}\\
\to y' = \dfrac{(x)'(1 - \cos x) - x(1-\cos x)'}{(1-\cos x)^2}\\
\to y' = \dfrac{1 - \cos x - x\sin x}{(1- \cos x)^2}\\
b)\quad y = \dfrac{1-\sin x}{1+\sin x}\\
\to y' = \dfrac{(1-\sin x)'(1+\sin x) - (1-\sin x)(1+\sin x)'}{(1+\sin x)^2}\\
\to y' = \dfrac{-\cos x(1 + \sin x) - (1-\sin x)\cos x}{(1+\sin x)^2}\\
\to y' = \dfrac{-\cos x - \sin x\cos x - \cos x+ \sin x\cos x}{(1+\sin x)^2}\\
\to y' = - \dfrac{2\cos x}{(1+\sin x)^2}\\
c)\quad y = \dfrac{\cos x}{\sin x + \cos x}\\
\to y' = \dfrac{(\cos x)'(\sin x + \cos x) - \cos x(\sin x + \cos x)'}{(\sin x + \cos x)^2}\\
\to y' = \dfrac{-\sin x(\sin x + \cos x) - \cos x(\cos x - \sin x)}{(\sin x + \cos x)^2}\\
\to y' = \dfrac{-\sin^2x - \sin x\cos x - \cos^2x + \sin x\cos x}{(\sin x + \cos x)^2}\\
\to y' = - \dfrac{1}{(\sin x + \cos x)^2}\\
d)\quad y = \dfrac{\sin x- \cos x}{\sin x + \cos x}\\
\to y' = \dfrac{(\sin x - \cos x)'(\sin x + \cos x) - (\sin x - \cos x)(\sin x + \cos x)'}{(\sin x + \cos x)^2}\\
\to y' = \dfrac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}\\
\to y' = \dfrac{\sin^2x + \cos^2x + 2\sin x\cos x + \sin^2x + \cos^2x - 2\sin x\cos x}{(\sin x + \cos x)^2}\\
\to y' = \dfrac{2}{(\sin x + \cos x)^2}\\
5)\\
a)\quad y = \sqrt{\tan^3x}\\
\to y' = \dfrac{(\tan^3x)'}{2\sqrt{\tan^3x}}\\
\to y' = \dfrac{3\tan^2.(\tan x)'}{2\sqrt{\tan^3x}}\\
\to y' = \dfrac{3\tan^2x}{2\cos^2x\sqrt{\tan^3x}}\\
\to y'= \dfrac{3\sqrt{\tan^3x}}{2\cos x.\tan x}\\
\to y' = \dfrac{3\sqrt{\tan^3x}}{\sin2x}\\
b)\quad y = \dfrac{2}{\cos\left(\dfrac{\pi}{6}-5x\right)}\\
\to y' = 2\cdot\left[- \dfrac{\cos\left(\dfrac{\pi}{6}-5x\right)'}{\cos^2\left(\dfrac{\pi}{6}-5x\right)}\right]\\
\to y' = -2\cdot\dfrac{\left(\dfrac{\pi}{6}-5x\right)'.\left[-\sin\left(\dfrac{\pi}{6}-5x\right)\right]}{\cos^2\left(\dfrac{\pi}{6}-5x\right)}\\
\to y'=-2\cdot \dfrac{-5.\left[-\sin\left(\dfrac{\pi}{6}-5x\right)\right]}{\cos^2\left(\dfrac{\pi}{6}-5x\right)}\\
\to y' = - \dfrac{10\sin\left(\dfrac{\pi}{6}-5x\right)}{\cos^2\left(\dfrac{\pi}{6}-5x\right)}\\
\end{array}\)