derivative of (x^4+3/(6x))

解

\frac{d}{d x} (\frac{x ^{4} + 3}{6 x})

\frac{x ^{4} - 1}{2 x ^{2}}

解答ステップ

\frac{d}{d x} (\frac{x ^{4} + 3}{6 x})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{6} \frac{d}{d x} (\frac{x ^{4} + 3}{x})

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( x ^{4} + 3 ) x - \frac{d x}{d x} ( x ^{4} + 3 )}{x ^{2}}

\frac{d}{d x} (x^{4} + 3) = 4 x^{3}

\frac{d x}{d x} = 1

= \frac{1}{6} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + 3 )}{x ^{2}}

簡素化

\frac{1}{6} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + 3 )}{x ^{2}} : \frac{x ^{4} - 1}{2 x ^{2}}

= \frac{x ^{4} - 1}{2 x ^{2}}