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Verified answer

А) 
f(x)=sin(2-3x)
f ' (x)= -3cos(2-3x)

б)
y=e^(cos²x)
y ' = e^(cos²x) * 2cosx*(-sinx)= -e^(cos²x)*sin2x=-sin2x*e^(cos²x)

4)
∛1,06 =(1+0,06)^(¹/₃)
f(x₀+Δx)≈f(x₀)+ f '(x₀)*Δx
x₀=1             Δx=0.06
f(x₀)=1^(¹/₃)=1
f '(x)=1/(3 ³√x²)
f '(1)=1/3

∛(1.06)=1+ ¹/₃ * 0.06=1+0.02=1.02

5.
y=cos(5x)
y ' = (cos(5x))' * (5x)' = -5sin5x
y '' = (-5sin5x)' = -5(sin(5x))' * (5x)' = -25cos5x