What is the d/(dt)((x)(2-x)(1-x)) ?

The d/(dt)((x)(2-x)(1-x)) is 3t^2-6t+2

What is the first d/(dt)((x)(2-x)(1-x)) ?

The first d/(dt)((x)(2-x)(1-x)) is 3t^2-6t+2

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\frac{d}{d t} ((t) (2 - t) (1 - t))

3 t^{2} - 6 t + 2

Solution steps

\frac{d}{d t} ((t) (2 - t) (1 - t))

Apply the Product Rule:

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

f = t, g = (2 - t) (1 - t)

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

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

\frac{d}{d t} ((2 - t) (1 - t)) = 2 t - 3

= 1 \cdot (2 - t) (1 - t) + (2 t - 3) t

Simplify

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

= 3 t^{2} - 6 t + 2

\frac{x ^{2} ( 3 y - 2 x )}{6}

\frac{( x ^{4} + 1 ) ^{2}}{x ^{4}}

\frac{( - 3 x + 12 )}{x ( x - 3 )}

(3 x^{2} + 2 x y + 3 y^{2}) d x

\frac{( b s + a ) ^{2}}{s ^{3}}

What is the d/(dt)((x)(2-x)(1-x)) ?
The d/(dt)((x)(2-x)(1-x)) is 3t^2-6t+2
What is the first d/(dt)((x)(2-x)(1-x)) ?
The first d/(dt)((x)(2-x)(1-x)) is 3t^2-6t+2