What is the d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) ?

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

What is the first d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) ?

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

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\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)

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(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)

Simplify

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

(\frac{1 + 1}{2 x})^{8}

\frac{( 2 c t ^{2} + 3 ) t}{2}

\frac{( x + 14 )}{( x - 4 ) ( x + 2 )}

\frac{1}{5 ( 2 - x )}

(x + y + 3 z = 8,) \cdot (2 x + 3 y + z = 12)

What is the d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) ?
The d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) is k(2s+2sk+5k+2)
What is the first d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) ?
The first d/(ds)((k)(s^2+2s+4+s^2k+5sk-9k)) is k(2s+2sk+5k+2)