What is the d/(dt)((t)(k(t-C))) ?

The d/(dt)((t)(k(t-C))) is k(2t-C)

What is the first d/(dt)((t)(k(t-C))) ?

The first d/(dt)((t)(k(t-C))) is k(2t-C)

Popular >

\frac{d}{d t} ((t) (k (t - C)))

k (2 t - C)

Solution steps

\frac{d}{d t} ((t) (k (t - C)))

Treat

k, C

as a constant

Take the constant out:

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

= k \frac{d}{d t} (t (t - C))

Apply the Product Rule:

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

f = t, g = t - C

= k (\frac{d t}{d t} (t - C) + \frac{d}{d t} (t - C) t)

\frac{d t}{d t} = 1

\frac{d}{d t} (t - C) = 1

= k (1 \cdot (t - C) + 1 \cdot t)

Simplify

k (1 \cdot (t - C) + 1 \cdot t) : k (2 t - C)

= k (2 t - C)

\frac{1}{( 1 - x ) ( b - x ) ( d - x )}

2 (x + Δ x) - 1

α (5, - 1, 2) + β (- 7, - 2, 11)

(5 y + 12 x yz - 10 + 6 x^{4}) - (12 +, + 4 x)

- \frac{2}{( 2 + h ) ^{2}}

What is the d/(dt)((t)(k(t-C))) ?
The d/(dt)((t)(k(t-C))) is k(2t-C)
What is the first d/(dt)((t)(k(t-C))) ?
The first d/(dt)((t)(k(t-C))) is k(2t-C)