derivative of y^3+xy^2+y+xcos(3x-2)

解

\frac{d}{d x} (y^{3} + x y^{2} + y + x cos (3 x) - 2)

2 x y \frac{d y}{d x} - 3 x sin (3 x) + cos (3 x) + 3 y^{2} \frac{d y}{d x} + y^{2} + \frac{d y}{d x}

解答ステップ

\frac{d}{d x} (y^{3} + x y^{2} + y + x cos (3 x) - 2)

和/差の法則を適用:

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

= \frac{d}{d x} (y^{3}) + \frac{d}{d x} (x y^{2}) + \frac{d y}{d x} + \frac{d}{d x} (x cos (3 x)) - \frac{d}{d x} (2)

\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}

\frac{d}{d x} (x y^{2}) = y^{2} + 2 x y \frac{d y}{d x}

\frac{d}{d x} (x cos (3 x)) = cos (3 x) - 3 x sin (3 x)

\frac{d}{d x} (2) = 0

= 3 y^{2} \frac{d y}{d x} + y^{2} + 2 x y \frac{d y}{d x} + \frac{d y}{d x} + cos (3 x) - 3 x sin (3 x) - 0

簡素化

= 2 x y \frac{d y}{d x} - 3 x sin (3 x) + cos (3 x) + 3 y^{2} \frac{d y}{d x} + y^{2} + \frac{d y}{d x}