derivative of (2x+|x|/(1+|x|))

解

\frac{d}{d x} (\frac{2 x + ∣ x ∣}{1 + ∣ x ∣})

\frac{x + 2 ∣ x ∣}{∣ x ∣ ( 1 + ∣ x ∣ ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{2 x + ∣ x ∣}{1 + ∣ x ∣})

商の法則を適用:

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

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

\frac{d}{d x} (2 x + ∣ x ∣) = 2 + \frac{x}{∣ x ∣}

\frac{d}{d x} (1 + ∣ x ∣) = \frac{x}{∣ x ∣}

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

簡素化

\frac{( 2 + \frac{x}{∣ x ∣} ) ( 1 + ∣ x ∣ ) - \frac{x}{∣ x ∣} ( 2 x + ∣ x ∣ )}{( 1 + ∣ x ∣ ) ^{2}} : \frac{x + 2 ∣ x ∣}{∣ x ∣ ( 1 + ∣ x ∣ ) ^{2}}

= \frac{x + 2 ∣ x ∣}{∣ x ∣ ( 1 + ∣ x ∣ ) ^{2}}