d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k))

解答

\frac{d}{d s} ((k) (s^{2} + 2 s + 4 + s^{2} k + 5 s k - 9 k))

k (2 s + 2 s k + 5 k + 2)

求解步骤

\frac{d}{d s} ((k) (s^{2} + 2 s + 4 + s^{2} k + 5 s k - 9 k))

将常数提出:

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

= k \frac{d}{d s} (s^{2} + 2 s + 4 + s^{2} k + 5 s k - 9 k)

使用微分加减法定则:

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

= k (\frac{d}{d s} (s^{2}) + \frac{d}{d s} (2 s) + \frac{d}{d s} (4) + \frac{d}{d s} (s^{2} k) + \frac{d}{d s} (5 s k) - \frac{d}{d s} (9 k))

\frac{d}{d s} (s^{2}) = 2 s

\frac{d}{d s} (2 s) = 2

\frac{d}{d s} (4) = 0

\frac{d}{d s} (s^{2} k) = 2 k s

\frac{d}{d s} (5 s k) = 5 k

\frac{d}{d s} (9 k) = 0

= k (2 s + 2 + 0 + 2 k s + 5 k - 0)

化简

= k (2 s + 2 s k + 5 k + 2)

e x p an d (\frac{1 + 1}{2 x})^{8}

e x p an d \frac{( 2 c t ^{2} + 3 ) t}{2}

e x p an d \frac{( x + 14 )}{( x - 4 ) ( x + 2 )}

e x p an d \frac{1}{5 ( 2 - x )}

e x p an d (x + y + 3 z = 8,) \cdot (2 x + 3 y + z = 12)