What is the d/(dt)((x)(k(x^2-5x+4))) ?

The d/(dt)((x)(k(x^2-5x+4))) is k(3x^2(dx)/(dt)-10x(dx)/(dt)+4(dx)/(dt))

What is the first d/(dt)((x)(k(x^2-5x+4))) ?

The first d/(dt)((x)(k(x^2-5x+4))) is k(3x^2(dx)/(dt)-10x(dx)/(dt)+4(dx)/(dt))

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\frac{d}{d t} ((x) (k (x^{2} - 5 x + 4)))

k (3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t})

Solution steps

\frac{d}{d t} ((x) (k (x^{2} - 5 x + 4)))

Take the constant out:

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

= k \frac{d}{d t} (x (x^{2} - 5 x + 4))

Apply the Product Rule:

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

f = x, g = x^{2} - 5 x + 4

= k (\frac{d x}{d t} (x^{2} - 5 x + 4) + \frac{d}{d t} (x^{2} - 5 x + 4) x)

\frac{d}{d t} (x^{2} - 5 x + 4) = 2 x \frac{d x}{d t} - 5 \frac{d x}{d t}

= k (\frac{d x}{d t} (x^{2} - 5 x + 4) + (2 x \frac{d x}{d t} - 5 \frac{d x}{d t}) x)

Expand

\frac{d x}{d t} (x^{2} - 5 x + 4) + (2 x \frac{d x}{d t} - 5 \frac{d x}{d t}) x : 3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t}

= k (3 x^{2} \frac{d x}{d t} - 10 x \frac{d x}{d t} + 4 \frac{d x}{d t})

, 3 (4 - 2 i)

\frac{d}{d t} ((x) (- 2 x^{2} + 16 x))

d x (6 x^{3} + 2 x + 1)

\frac{s - 1}{s ^{2} ( s - 2 ) ( s + 1 )}

What is the d/(dt)((x)(k(x^2-5x+4))) ?
The d/(dt)((x)(k(x^2-5x+4))) is k(3x^2(dx)/(dt)-10x(dx)/(dt)+4(dx)/(dt))
What is the first d/(dt)((x)(k(x^2-5x+4))) ?
The first d/(dt)((x)(k(x^2-5x+4))) is k(3x^2(dx)/(dt)-10x(dx)/(dt)+4(dx)/(dt))