(-x^2sin(x)+2pixsin(x)-2picos(x)+2pi)^'

解

(- x^{2} sin (x) + 2 π x sin (x) - 2 π cos (x) + 2 π)^{^{'}}

- x^{2} cos (x) + 2 π x cos (x) - 2 x sin (x) + 4 π sin (x)

解答ステップ

(- x^{2} sin (x) + 2 π x sin (x) - 2 π cos (x) + 2 π)^{'}

和/差の法則を適用:

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

= - (x^{2} sin (x))^{^{'}} + (2 π x sin (x))^{^{'}} - (2 π cos (x))^{^{'}} + (2 π)^{^{'}}

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

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

(2 π cos (x))^{'} = - 2 π sin (x)

(2 π)^{'} = 0

= - (2 x sin (x) + cos (x) x^{2}) + 2 π (sin (x) + x cos (x)) - (- 2 π sin (x)) + 0

簡素化

- (2 x sin (x) + cos (x) x^{2}) + 2 π (sin (x) + x cos (x)) - (- 2 π sin (x)) + 0 : - x^{2} cos (x) + 2 π x cos (x) - 2 x sin (x) + 4 π sin (x)

= - x^{2} cos (x) + 2 π x cos (x) - 2 x sin (x) + 4 π sin (x)