derivative of 2y-1/x+cos(3x)

解

\frac{d}{d x} (2 y - \frac{1}{x} + cos (3 x))

2 \frac{d y}{d x} + \frac{1}{x ^{2}} - 3 sin (3 x)

解答ステップ

\frac{d}{d x} (2 y - \frac{1}{x} + cos (3 x))

和/差の法則を適用:

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

= \frac{d}{d x} (2 y) - \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (cos (3 x))

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

\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

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

= 2 \frac{d y}{d x} - (- \frac{1}{x ^{2}}) - sin (3 x) \cdot 3

簡素化

= 2 \frac{d y}{d x} + \frac{1}{x ^{2}} - 3 sin (3 x)