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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

f = x, g = t^{2} - t

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

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

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

\frac{( ( 64 x + 32 ) ( 2 x + 1 ) )}{2}

(- 1, - π, \frac{1}{2 π}) \cdot (- π, - 2, 0)

2 (6 e + 2)

\frac{1}{( p ( a - p ) )}

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

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