d/(ds)((k)(-s(s+1)(s+2)))

解答

\frac{d}{d s} ((k) (- s (s + 1) (s + 2)))

- k (3 s^{2} + 6 s + 2)

求解步骤

\frac{d}{d s} ((k) (- s (s + 1) (s + 2)))

将常数提出:

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

= k \frac{d}{d s} (- s (s + 1) (s + 2))

将常数提出:

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

= k (- \frac{d}{d s} (s (s + 1) (s + 2)))

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = s, g = (s + 1) (s + 2)

= k (- (\frac{d s}{d s} (s + 1) (s + 2) + \frac{d}{d s} ((s + 1) (s + 2)) s))

\frac{d s}{d s} = 1

\frac{d}{d s} ((s + 1) (s + 2)) = 2 s + 3

= k (- (1 \cdot (s + 1) (s + 2) + (2 s + 3) s))

化简

k (- (1 \cdot (s + 1) (s + 2) + (2 s + 3) s)) : - k (3 s^{2} + 6 s + 2)

= - k (3 s^{2} + 6 s + 2)

e x p an d \frac{k ( s + 4 )}{s ^{2} + 4 s + 4}

e x p an d - \frac{16 ln ( 3 )}{35 ( 1 - e ^{\frac{2 l n ( 3 )}{35}} )}

e x p an d 6 x^{2} + 19 + 9 x^{2} - 8

e x p an d V (3, - 1) F (3, 1)

e x p an d (y + ln ∣ e^{y} - 2 ∣) \cdot ln (x)