Question

Здравствуйте. Помогите пожалуйста

Answer 1

$y=(3x+1)^3\cdot cos^3(x^2+2x+1)+\pi ^3\; \; ,\; \; x_0=-1\\\\\\y'=\Big ((3x+1)^3\Big )'\cdot cos^3(x^2+2x+1)+(3x+1)^3\cdot \Big (cos^3(x^2+2x+1)\Big )'+(\pi ^3)'=\\\\=3\cdot (3x+1)^2\cdot 3\cdot cos^3(x^2+2x+1)+\\\\+(3x+1)^3\cdot 3cos^2(x^2+2x+1)\cdot (-sin(x^2+2x+1))\cdot (2x+2)+0=\\\\=9(3x+1)^2\cdot cos^3(x+1)^2-6(3x+1)^3\cdot cos^2(x+1)^2\cdot sin(x+1)^2\cdot (x+1)\\\\\\y'(-1)=9\cdot 4\cdot cos^30-6\cdot (-8)\cdot cos^20\cdot sin0\cdot 0=9\cdot 4\cdot 1=36\\\\\\P.S.\; \; x^2+2x+1=(x+1)^2$

Answer 2

$y=(3x+1)^{3} *Cos^{3}(x^{2}+2x+1)+\pi^{3}\\\\y'=[(3x+1)^{3}]'*Cos^{3}(x^{2}+2x+1)+(3x+1)^{3}*[Cos^{3} (x^{2}+2x+1)]'+(\pi ^{3})'=3(3x+1)^{2}*(3x+1)'*Cos^{3}(x^{2}+2x+1)+(3x+1)^{3}*3Cos^{2}(x^{2}+2x+1)*[Cos(x^{2}+2x+1)]'*(x^{2}+2x+1)'+0=9(3x+1)^{2}*Cos^{3}(x^{2}+2x+1)+(3x+1)^{3}*3Cos^{2}(x^{2}+2x+1)*[(-Sin(x^{2}+2x+1)]*(2x+2)=9(3x+1)^{2}*Cos^{3}(x^{2}+2x+1)-(3x+1)^{3}*(6x+6)*Cos^{2}(x^{2}+2x+1)*Sin(x^{2}+2x+1)$

$y'(x_{o})=y'(-1)= 9*(3*(-1)+1)^{2}*Cos^{3}((-1)^{2}+2*(-1)+1)-(3*(-1)+1)^{3}*(6*(-1)+6)*Cos^{2}((-1)^{2}+2*(-1)+1)*Sin((-1)^{2}+2*(-1)+1)=9*4*Cos^{3}0-(-8)*0*Cos^{2}0*Sin0=9*4*1-0=36\\\\Otvet:\boxed{36}$