((1+sin(x))^2+(1+cos(x))^4)^'

解

((1 + sin (x))^{2} + (1 + cos (x))^{4})^{^{'}}

- 5 sin (2 x) + 2 cos (x) - 4 cos^{3} (x) sin (x) - 12 cos^{2} (x) sin (x) - 4 sin (x)

解答ステップ

((1 + sin (x))^{2} + (1 + cos (x))^{4})^{'}

簡素化

(1 + sin (x))^{2} + (1 + cos (x))^{4} : cos^{4} (x) + 4 cos^{3} (x) + sin^{2} (x) + 6 cos^{2} (x) + 2 sin (x) + 4 cos (x) + 2

= (cos^{4} (x) + 4 cos^{3} (x) + sin^{2} (x) + 6 cos^{2} (x) + 2 sin (x) + 4 cos (x) + 2)^{^{'}}

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (cos^{4} (x))^{^{'}} + (4 cos^{3} (x))^{^{'}} + (sin^{2} (x))^{^{'}} + (6 cos^{2} (x))^{^{'}} + (2 sin (x))^{^{'}} + (4 cos (x))^{^{'}} + 2^{^{'}}

(cos^{4} (x))^{'} = - 4 cos^{3} (x) sin (x)

(4 cos^{3} (x))^{'} = - 12 cos^{2} (x) sin (x)

(sin^{2} (x))^{'} = sin (2 x)

(6 cos^{2} (x))^{'} = - 12 cos (x) sin (x)

(2 sin (x))^{'} = 2 cos (x)

(4 cos (x))^{'} = - 4 sin (x)

2^{'} = 0

= - 4 cos^{3} (x) sin (x) - 12 cos^{2} (x) sin (x) + sin (2 x) - 12 cos (x) sin (x) + 2 cos (x) - 4 sin (x) + 0

簡素化

- 4 cos^{3} (x) sin (x) - 12 cos^{2} (x) sin (x) + sin (2 x) - 12 cos (x) sin (x) + 2 cos (x) - 4 sin (x) + 0 : - 5 sin (2 x) + 2 cos (x) - 4 cos^{3} (x) sin (x) - 12 cos^{2} (x) sin (x) - 4 sin (x)

= - 5 sin (2 x) + 2 cos (x) - 4 cos^{3} (x) sin (x) - 12 cos^{2} (x) sin (x) - 4 sin (x)