y = (3x+1)^3 * cos^3(x^2 + 2x +1) + pi^3 х0 = - 1
y'= ( (3x+1)^3 )' * cos^3(x^2 + 2x +1) + (3x+1)^3 * (cos^3(x^2 + 2x +1))' =
= 3*(3x+1)^2 *3 *cos^3(x^2 + 2x +1) - (3x+1)^3 *3 cos^2(x^2 + 2x +1)* *Sin(x^2+2x+1)*(2x +2)=
=9(3x+1)^2 *cos^3(x^2 + 2x +1) -3(2x+2)(3x+1)^3Cos^2(x^2 + 2x +1)* *Sin(x^2+2x+1)
при х = -1 получим:
y' = 9*(-2)*Cos^3(0) -3*0*(-2)*Cos0*Sin0= -18
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