derivative of |x-1|+|2x+3|

解答

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

\frac{x - 1}{∣ x - 1 ∣} + \frac{2 ( 2 x + 3 )}{∣ 2 x + 3 ∣}

求解步骤

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

使用微分加减法定则:

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

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

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

\frac{d}{d x} (∣ 2 x + 3 ∣) = \frac{2 ( 2 x + 3 )}{∣ 2 x + 3 ∣}

= \frac{x - 1}{∣ x - 1 ∣} + \frac{2 ( 2 x + 3 )}{∣ 2 x + 3 ∣}

a re a y = 2 x^{2}, y = 8 x - 6 x^{2}

\int 5 - cos (x) d x

d er i v a t i v e 6 u

x \to \frac{π}{2} lim (sec^{2} (x))

\int x \cdot 1 + \frac{1}{4 x} d x