What is the d/(dt)((x)(t^2-4t)) ?

The d/(dt)((x)(t^2-4t)) is (dx)/(dt)(t^2-4t)+(2t-4)x

What is the first d/(dt)((x)(t^2-4t)) ?

The first d/(dt)((x)(t^2-4t)) is (dx)/(dt)(t^2-4t)+(2t-4)x

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

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

Solution steps

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

Apply the Product Rule:

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

f = x, g = t^{2} - 4 t

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

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

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

\frac{2 π x ^{2} - 44 x ^{2} + 1440 x}{9}

(0, - 2) y (0, - 2)

\frac{5 ( 9 - 21 )}{3}

\frac{2 e ^{x} + 5}{in ( 3 + x )}

\frac{( s ^{2} - 4 )}{( s ^{2} + 4 ) ^{2}}

What is the d/(dt)((x)(t^2-4t)) ?
The d/(dt)((x)(t^2-4t)) is (dx)/(dt)(t^2-4t)+(2t-4)x
What is the first d/(dt)((x)(t^2-4t)) ?
The first d/(dt)((x)(t^2-4t)) is (dx)/(dt)(t^2-4t)+(2t-4)x