Question

Найдите f"(x) вторая производная функция

Answer 1

$5)f(x)=(x+1)^{2} Cos2x\\\\f'(x)=[(x+1)^{2}]'*Cos2x+(x+1)^{2}*(Cos2x)'= \\\\=2(x+1)*Cos2x-(x+1)^{2}*2Sin2x=2[(x+1)Cos2x-(x+1)^{2}Sin2x)]\\\\f''(x)=2[(x+1)'Cos2x+(x+1)*(Cos2x)'-[(x+1)^{2}]'Sin2x-(x+1)^{2}(Sin2x)']=\\\\=2[Cos2x-2(x+1)Sin2x-2(x+1)Sin2x-(x+1)^{2}*2Cos2x]=\\\\=2(Cos2x-4(x+1)Sin2x-2(x+1)^{2}Cos2x)$

$6)f(x)=xCos(x^{2}+1)\\\\f'(x)=x'*Cos(x^{2}+1)+x*(Cos(x^{2}+1))'=Cos(x^{2}+1)-x*2xSin(x^{2}+1)=\\\\=Cos(x^{2}+1)-2x^{2} Sin(x^{2}+1)\\\\f''(x)=[Cos(x^{2}+1)]'- 2[(x^{2})'*Sin(x^{2}+1)+x^{2}*(Sin(x^{2}+1))'] =\\\\=-2xSin(x^{2}+1)-2(2xSin(x^{2}+1)+2x^{3}Cos(x^{2}+1)=\\\\=-2xSin(x^{2}+1)-4xSin(x^{2}+1)-4x^{3}Cos(x^{2}+1)=\\\\ -6xSin(x^{2}+1)-4x^{3}Cos(x^{2}+1)$

Answer 2

5)

$f'(x) = 2(x + 1) \times \cos(2x) + ( - \sin(2x) ) \times 2 \times {(x + 1)}^{2} = \\ =2 (x + 1) \times ( \cos(2x) - (x + 1) \sin(2x) )$

$f''(x) = 2( \cos(2x) - (x + 1) \sin(2x) ) + 2(x + 1)( - 2 \sin(2x) - \sin(2x) - (x + 1) \times 2 \cos(2x) ) = \\ = 2 \cos(2x) - 2x \sin(2x) - 2 \sin(2x) - 4x \sin(2x) - 4 \sin(2x) - 2 \sin(2x) - 4x \cos(2x) - 4 \cos(2x) = \\ = - 2 \cos(2x) - 6 \sin(2x) - 6x \sin(2x) - 4x \cos(2x)$

6)

$f'(x) = 1 \times \cos( {x}^{2} + 1 ) - \sin( {x}^{2} + 1) \times 2x \times x = \\ = \cos( {x}^{2} + 1) - 2 {x}^{2} \sin( {x}^{2} + 1)$

$f''(x) = - 2x \sin( {x}^{2} + 1) - 4x \sin( {x}^{2} +1 ) - 2 {x}^{2} \times \cos( {x}^{2} + 1) \times 2x = 2x \sin( {x}^{2} + 1 ) - 4 {x}^{3} \cos( {x}^{2} + 1 )$